On Convolution Squares of Singular Measures
On Convolution Squares of Singular Measures
by
Vincent Chan
A thesis presented to the University of Waterloo in fulfillment of the thesis requirement for the degree of Master of Mathematics in Pure Mathematics
Waterloo, Ontario, Canada, 2010
⃝c Vincent Chan 2010
I hereby declare that I am the sole author of this thesis. This is a true copy of the thesis, including any required final revisions, as accepted by my examiners.
I understand that my thesis may be made electronically available to the public.
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Abstract
We prove that if 1 > > 1/2, then there exists a probability measure such that the Hausdorff dimension of its support is and ∗ is a Lipschitz function of class − 1/2.
v
Acknowledgements
I would like to deeply thank my supervisor, Kathryn Hare, for her excellent supervision; her comments were useful and insightful, and led not only to finding and fixing mistakes, but also to a more elegant way of presenting various arguments. A careful comparison of earlier drafts of this document and its current version would reveal the subtle errors detected by her keen eye.
This thesis would not exist at all if it were not for Professor Thomas K¨orner,whom I thank for his help in clarifying points in his paper, which is the central focus in the pages to come.
I am indebted to my friends and colleagues for their support and advice, particularly that of Faisal Al-Faisal, who, being in a similar situation to myself (including the bizarre sleep schedule), has become comparable to a brother in arms.
Last but certainly not least, I am grateful for the support of my fianc´eeVicki Nguyen, who, despite having no formal education in pure mathematics, suffered through reading dozens of pages while proofreading. She has never failed to provide encouragement when it was required most, and surely has kept me sane throughout the process.
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Contents
1 Preliminaries1 1.1 Introduction...... 1 1.2 A tightness condition...... 7
2 A Probabilistic Result 11
3 A Key Lemma 21
4 The Main Theorem 35 4.1 -Lipschitz functions...... 35 4.2 Density results...... 41
APPENDICES 57
A Measure Theory 59
B Convergence of Fourier Series 63
C Probability Theory 65
D The Hausdorff Metric 67
Bibliography 74
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Chapter 1
Preliminaries
Unless otherwise stated, we will be working on T = ℝ/ℤ = [−1/2, 1/2], and only with positive, Borel, regular measures. We will use to denote the Lebesgue measure on T (and dx, dy, dt, and so forth to denote integration with respect to Lebesgue measure), and ∣E∣ = (E). Measures that are absolutely continuous or singular are taken with respect to in the case of T, and (left) Haar measure, should we require a more general setting. When speaking of measures living in an Lp space or having other properties associated to functions, we mean the measure is absolutely continuous and we consider its Radon- Nikodym derivative to have this property.
1.1 Introduction
The absolutely continuous measures form an ideal in the algebra of measures, so that for an absolutely continuous measure , ∗ is still absolutely continuous and so the Radon- Nikodym theorem gives rise to an associated function f ∈ L1 such that ∗ = f. For any -finite measure , Lebesgue’s Decomposition theorem decomposes as = ac + s, where ac is absolutely continuous and s is singular. It is thus natural to ask whether a similar situation holds with singular measures, that is, given a singular measure , does there exist a function f ∈ L1 with ∗ = f? This is easily seen to be false, for if
1 P P = n anxn is a discrete measure, then ∗ = n,m anamxn+xm is also a discrete measure.
To progress further, it may be helpful to appeal to an extension of Lebesgue’s Decom-
position theorem, which gives for any -finite measure three unique measures ac, cs
and d which are absolutely continuous, continuous singular, and discrete, respectively,
such that = ac + cs + d. Then we ask if we might recover the nice property of ∗ being absolutely continuous in the case of being a continuous singular measure. This is still not always the case, as there exist continuous singular measures on T where even their nth convolution power is singular for any n; for such an example we will need a preliminary definition.
Definition 1.1. Let −1/2 ≤ ak ≤ 1/2 and define the trigonometric polynomials on T
N Y k PN (t) = (1 + 2ak cos 3 t). k=1
It is a standard result that PN are probability measures which converge weak-∗ in M(T) to a probability measure with ⎧ 1 if n = 0, ⎨ M M (n) = Q P kj b j=1 akj if n = j=1 j3 , j = ±1, ⎩0 otherwise.
This measure is called the Riesz product measure associated to {ak}.
Lemma 1.2. If 1 and 2 are Riesz product measures, then so is 1 ∗ 2. Moreover, if i
is associated to {ak,i}, then 1 ∗ 2 is associated to {ak,1ak,2}.
Proof. We calculate the Fourier coefficients of the convolution: ⎧ 1 if n = 0, ⎨ M M k 1 ∗ 2(n) = 1(n)2(n) = Q P j (1.1) \ b b j=1 akj ,1akj ,2 if n = j=1 j3 , j = ±1, ⎩0 otherwise.
2 Then we can define the trigonometric polynomials on T
N Y k PN (t) = (1 + 2ak,1ak,2 cos 3 t), k=1 noticing that −1/2 ≤ −1/4 ≤ ak,1ak,2 ≤ 1/4 ≤ 1/2. The weak-∗ limit of PN produces a Riesz product measure whose coefficients satisfy (1.1), so we are finished by uniqueness.
A standard result (see, for example, [6] Theorem 7.2.2) gives that a Riesz product P∞ 2 measure is singular if and only if k=1 ak = ∞. In particular, suppose is a Riesz product measure associated to the constant sequence {c} for some constant 0 < c ≤ 1/2. P∞ 2 n Certainly k=1 c = ∞, so that is singular. Inductively applying Lemma 1.2, = ∗ ∗ ⋅ ⋅ ⋅ ∗ (the nth convolution power of ) is a Riesz product measure associated to n P∞ 2n n {c }. Then k=1 c = ∞, and so remains singular. It is also not hard to show that Riesz product measures are continuous, in general.
It should be remarked that the choice of frequencies 3k is not essential, and Riesz product measures may be generalized further. In addition, the concept of Riesz product measures as a whole and the conclusion that there exist continuous singular measures whose convolution powers always remain singular can be extended to any infinite, compact, abelian group (see [6]).
We now know it is impossible to hope the square convolution of a singular measure is absolutely continuous in general, even when restricting to the case of continuous singular measures. However, we may expect that are still some positive results; convolution behaves a smoothing operation, evidenced by the fact that both the continuous measures and the absolutely continuous measures form ideals in the space of measures. Indeed, a famous result due to Wiener and Wintner [16] produces a singular measure on T such that −1/2+" b(n) = o(∣n∣ ) as n → ∞ for every " > 0. By an application of the Hausdorff-Young inequality, such a measure can be shown to have the property that its square convolution ∗ is absolutely continuous, and in fact, ∗ ∈ Lp(T) for every p ≥ 1. Other authors have generalized this result; Hewitt and Zuckerman [9] constructed a singular measure on any non-discrete, locally compact, abelian group G such that ∗ ∈ Lp(G) for all p ≥ 1. It is worth noting that the condition of non-discrete cannot be
3 dropped, since otherwise all measures would be absolutely continuous. Using Rademacher- Riesz products, Karanikas and Koumandos [10] prove the existence of a singular measure with ∗ ∈ L1(G) for any non-discrete, locally compact group G. Shortly thereafter, Dooley and Gupta [3] produced a measure with ∗ ∈ Lp(G) for every p ≥ 1 in the case of compact, connected groups and compact Lie groups using the theory of compact Lie groups.
Saeki [15] took this concept further in a different direction by proving the existence of a singular measure on T with support having zero Lebesgue measure such that ∗ has a uniformly convergent Fourier series. This is an improvement on previous results; in this case the continuity of the partial Fourier sums for ∗ ensure that their limit, namely ∗ , is continuous. Then ∗ ∈ L∞(T) and subsequently ∗ ∈ Lp(T) for p ≥ 1 since T is compact. Generalizing this, Gupta and Hare [7] show such a measure exists (now using Haar measure in place of Lebesgue measure) when replacing T with any compact, connected group.
K¨orner[11] recently expanded on Saeki’s work; to discuss how, we need the following definitions.
Definition 1.3. For 1 ≥ ≥ 0, we say a function f : T → ℂ is Lipschitz of class 1, or simply -Lipschitz and write f ∈ Λ if
sup sup ∣ℎ∣− ∣f(t + ℎ) − f(t)∣ < ∞. (1.2) t∈T ℎ∕=0
Some references define a function to be Lipschitz of class if there is a constant C with ∣f(x) − f(y)∣ ≤ C∣x − y∣ for every x, y ∈ T. These definitions are in fact equivalent, with a straightforward proof.
Lemma 1.4. f ∈ Λ if and only if ∣f(x) − f(y)∣ ≤ C∣x − y∣ for a constant C, for every x, y ∈ T.
Proof. This is a simple matter of relabeling; use t = y and ℎ = x − y.
1Functions with this property are also sometimes referred to as H¨olderof class .
4 It is useful to note that the Λ classes are nested downwards:
Lemma 1.5. Suppose 1 ≥ i ≥ 0 for i = 1, 2. If 1 ≤ 2, then Λ 1 ⊇ Λ 2 .
− 1 − 2 Proof. Since 1 ≥ i ≥ 0, for ℎ ∈ T = [−1/2, 1/2] we get that ∣ℎ∣ ≤ ∣ℎ∣ . Thus if
sup sup ∣ℎ∣− 2 ∣f(t + ℎ) − f(t)∣ < ∞, t∈T ℎ∕=0 we certainly have sup sup ∣ℎ∣− 1 ∣f(t + ℎ) − f(t)∣ < ∞, t∈T ℎ∕=0 implying Λ 1 ⊇ Λ 2 .
We will also require the notion of Hausdorff dimension, which we recall here:
Definition 1.6. Fix ≥ 0 and let > 0. For a set E, define
( ∞ ∞ ) X [ ℋ (E) = inf ∣Ei∣ : Ei ⊇ E, ∣Ei∣ ≤ , i=1 i=1 where Ei can be taken to be intervals. Notice ℋ (E) is monotone increasing as decreases, since fewer permissible collections of sets are taken in the infimum. Then we take
ℋ (E) = lim ℋ (E) = sup ℋ (E), →0+ >0 and we call ℋ (E) the -Hausdorff measure of E; it is standard that this is indeed a measure (see [4], for example). Of special interest is the case = 1, in which case we recover the usual Lebesgue measure. It is easy to show that there is a critical value for such that ℋ (E) = ∞ for less than this critical value, and ℋ (E) = 0 for greater than this critical value. Then we define the Hausdorff dimension of E by
dimH (E) = sup{ : ℋ (E) > 0} = sup{ : ℋ (E) = ∞} = inf{ : ℋ (E) < ∞} = inf{ : ℋ (E) = 0}.
Heuristically, the Hausdorff dimension allows us to compare sets that are too sparse for the Lebesgue measure to be of use.
5 For our results, we will need to have a way of describing where a measure “lives”; we make this formal by introducing the concept of the support of a measure.
Definition 1.7. Let (X,T ) be a topological space, and be a Borel measure on (X,T ). Then the support of is defined to be the set of all points x in X for which every open neighborhood Nx of x has positive measure:
supp = {x ∈ X : (Nx) > 0, ∀x ∈ Nx ∈ T }. (1.3)
Among other things, K¨ornerdemonstrated the existence of a probability measure whose support has a prescribed Hausdorff dimension (between 1 and 1/2) such that ∗ is a Lipschitz function. Our goal will be to prove one of his main results, which is the following:
Theorem 4.19. If 1 > > 1/2, then there exists a probability measure such that dimH (supp ) = and ∗ = f where f ∈ Λ −1/2.
We will show that this theorem is indeed an extension of Saeki’s work. Using Lemma
1.4, we can see that f ∈ Λ yields a constant C such that
∣f(x + ℎ) − f(x)∣ ≤ C∣ℎ∣ for any x and ℎ, implying f(x + ℎ) − f(x) = o(((ln ∣ℎ∣−1)−1) (see Appendix B.1). By the Dini-Lipschitz test (see Appendix B.2), f has a uniformly convergent Fourier series. This says that the measure K¨ornerproduced has the same property as Saeki’s, provided it were also singular. This fact will hold due to the Hausdorff dimension condition; indeed, the fact that dimH (supp ) < 1 implies that ∣supp ∣ = 0 by definition of the dimension and recalling ℋ1 is simply Lebesgue measure. Since we always have ((supp )c) = 0, this provides the necessary decomposition for to be singular.
Consider now the condition 1 > > 1/2. As we have seen, this ensures the constructed measure will be singular since sets of Hausdorff dimension less than 1 have zero Lebesgue measure; however the converse is not true. Indeed, it is well known that there exists a set of Hausdorff dimension 1 with zero Lebesgue measure. Then the case = 1 could give still
6 give rise to singular measures. Since we are working on T which has Hausdorff dimension 1, monotonicity of Hausdorff dimension (easily seen from monotonicity of the Hausdorff measure) ensures we need not consider the case > 1. We shall see soon that we also need ≥ 1/2 for a positive result involving this Lipschitz condition, after proving a tightness condition.
1.2 A tightness condition
In this section, we aim to show that the result in Theorem 4.19 is nearly the best we can hope for. In particular, we will show that if the support of a measure has Hausdorff measure dimension and ∗ ∈ Λ , then −1/2 ≥ . To begin, we mention a relationship between the Hausdorff dimension of the support of a measure and its Fourier coefficients.
Definition 1.8. We define the s-energy of a finite, compactly supported measure on ℝn, denoted Is(), to be ZZ d(x) d(y) I () = . s ∣x − y∣s
Lemma 1.9. Suppose is a non-zero measure, 0 < < 1, and
X ∣∣2 b < ∞. ∣k∣1− k∕=0
Then dimH (supp ) ≥ .
Proof. By Theorem 2.2 in [8], with d = 1 and 0 < < 1, we have for some constant b that ! X ∣(k)∣2 I () ≤ b ∣(0)∣2 + b < ∞. b ∣k∣1− k∕=0
Then by Theorem 4.13 in [5], dimH (supp ) ≥ .
Now, we develop a bound for portions of the ℓp norm of the Fourier coefficients of a -Lipschitz function.
7 Lemma 1.10. If f ∈ Λ and 0 < p ≤ 2, then
1 X p 1−p( + ) ∣fb(k)∣ ≤ C1n 2 , n≤∣k∣≤2n−1 for some constant C1.
Proof. Fix ℎ ∈ [−1/2, 1/2]. Let g(t) = f(t + ℎ) − f(t − ℎ). A simple calculation shows that the Fourier coefficients of g are given by
gb(k) = 2i sin(2kℎ)fb(k).
P 2 2 By Parseval’s identity, k ∣gb(k)∣ = ∥g∥ , so that X Z 4 ∣fb(k)∣2 sin2(2kℎ) = ∣f(t + ℎ) − f(t − ℎ)∣2 dt. k
Since f ∈ Λ , there is a constant C such that
∣f(t + ℎ) − f(t − ℎ)∣ ≤ C∣(t + ℎ) − (t − ℎ)∣ = C∣2ℎ∣ , for all t ∈ T, so X Z 4 ∣fb(k)∣2 sin2(2kℎ) ≤ (C∣2ℎ∣ )2 dt = 4 C2∣ℎ∣2 ≤ 4C2∣ℎ∣2 . k Thus for any n, X ∣fb(k)∣2 sin2(2kℎ) ≤ C2∣ℎ∣2 . n≤∣k∣≤2n−1
2 k As this holds for any ℎ ∈ [−1/2, 1/2], consider ℎ = 1/(8n). Notice sin ( 4n ) ≥ 1/2 for n ≤ ∣k∣ ≤ 2n − 1, so
X 2 2 2 2 1 2 −2 2 1 2 −2 ∣fb(k)∣ ≤ 2C ∣ℎ∣ = 2C ( 8 ) n ≤ 2C ( 8 ) n . (1.4) n≤∣k∣≤2n−1
2 1 2 1 2 If p = 2, we are done; taking C1 = 2C ( 8 ) = 32 C , and (1.4) gives
1 X 2 −2 1−2( + ) ∣fb(k)∣ ≤ C1n = C1n 2 . n≤∣k∣≤2n−1
8 Otherwise, 0 < p < 2 and so 1 < 2/p, 1/(1 − p/2) < ∞; notice they are conjugate indices. p By H¨older’sinequality with counting measure on {∣fb(k)∣ }n≤∣k∣≤2n−1 and {1}n≤∣k∣≤2n−1, we have ⎛ ⎞p/2 ⎛ ⎞1−p/2 X p X p 2/p X 1/(1−p/2) ∣fb(k)∣ ≤ ⎝ (∣fb(k)∣ ) ⎠ ⎝ (1) ⎠ n≤∣k∣≤2n−1 n≤∣k∣≤2n−1 n≤∣k∣≤2n−1 ⎛ ⎞p/2 X 2 1−p/2 = ⎝ ∣fb(k)∣ ⎠ (2n) n≤∣k∣≤2n−1 p 1− 2 1 2 −2 p/2 2 ≤ (2C ( 8 ) n ) (2n) by (1.4) 1−p( +1/2) = C1n
C p as required, where C1 = 2( 8 ) is a constant.
Lemma 1.11. If is a measure with dimH (supp ) = and ∗ = f where f ∈ Λ , then − 1/2 ≥ .
Proof. Since f ∈ Λ , taking p = 1 in Lemma 1.10 yields
X 1−1( +1/2) (1−2 )/2 ∣fb(k)∣ ≤ C1n = C1n , n≤∣k∣≤2n−1
2 for some constant C1 depending on f. Since ∣fb(k)∣ = ∣b(k)∣ , we have
X 2 1−1( +1/2) (1−2 )/2 ∣b(k)∣ ≤ C1n = C1n , n≤∣k∣≤2n−1 and so if > 0, we have
2 2 (1−2 )/2 X ∣(k)∣ X ∣(k)∣ C1n b ≤ b ≤ = C n−(1+2 −2)/2 ∣k∣1− n1− n1− 1 n≤∣k∣≤2n−1 n≤∣k∣≤2n−1 for all n ≥ 1. In particular, for n = 2j for each j ≥ 0,
X ∣(k)∣2 b ≤ C (2j)−(1+2 −2)/2, ∣k∣1− 1 2j ≤∣k∣≤2j+1−1
9 so that ∞ ⎛ ⎞ ∞ X ∣(k)∣2 X X ∣(k)∣2 X b = b ≤ C (2j)−(1+2 −2)/2. ∣k∣1− ⎝ ∣k∣1− ⎠ 1 k∕=0 j=0 2j ≤∣k∣≤2j+1−1 j=0 Notice this converges whenever −(1 + 2 − 2)/2 < 0, that is, whenever (1 + 2 )/2 > .
Then by Lemma 1.9, dimH (supp ) ≥ , for every that satisfies (1 + 2 )/2 > . Thus dimH (supp ) ≥ (1 + 2 )/2, that is, − 1/2 ≥ as required.
Lemma 1.11 now gives us a motivation for the lower bound on ; taking = 0 we see that for ∗ ∈ Λ0, we must have dimH (supp ) ≥ 1/2. Of course, this still leaves open the question of what happens in the case of = 1/2; the boundary cases = 1 and = 1/2 are beyond the scope of this paper, as the proof relies on the strict inequalities.
In Chapter2, we prove a key lemma that provides us with a discrete measure satisfying useful boundedness properties which we help in our construction. The key to this lemma will be using a probabilistic argument, which will span most of the chapter.
In Chapter3, we utilize the key lemma from the previous chapter to construct an infinitely differentiable periodic function satisfying several key properties to be used in the main theorem.
In Chapter4, the main theorem will be proved, first by developing some complete metric spaces and applying the key lemma from the previous chapter in a density argument. Employing the Baire Category theorem will bridge the final gap towards the proof of the main theorem.
10 Chapter 2
A Probabilistic Result
The key result that we will need is not easy to prove directly, so we will prove a probabilistic variant of it instead, which will imply the result we need. We proceed by proving the following result:
Lemma 2.1. Suppose that 0 < Np ≤ 1 and m ≥ 2. Then if Y1,Y2,...,YN are independent random variables with
P(Yj = 1) = p, P(Yj = 0) = 1 − p, it follows that N ! X 2(Np)m P Y ≥ m ≤ . j m! j=1
Proof. First note that Yj ∈ {0, 1} almost surely, so that their sum cannot exceed N with positive probability. That is, we may assume without loss of generality that m ≤ N. For 1 ≤ k ≤ N, define ! N k N! k uk = p = p . k k!(N − k)! Since 0 < Np ≤ 1 and k, p > 0, we have (N − k)p = Np − kp ≤ 1. Then as k ≥ 1,
N! u pk+1 (N − k)p 1 1 k+1 = (k+1)!(N−k−1)! = ≤ ≤ . u N! k k + 1 k + 1 2 k k!(N−k)! p
11 Since Yj takes on only 0 or 1 almost surely,
N ! N N ! X X X P Yj ≥ m = P Yj = k j=1 k=m j=1 N ! X N k N−k = p (1 − p) since Yj are independent k=m k N ! N X N k X ≤ p = uk k=m k k=m N X um uk+1 1 ≤ k−m ≤ 2um since ≤ 2 uk 2 k=m N! 2(Np)m = 2 pm ≤ . m!(N − m)! m!
Definition 2.2. For a set A ⊆ ℝ and a random variable X, we define the random variable
X (A) by ⎧ ⎨1 if X(x) ∈ A, X (A)(x) = ⎩0 otherwise.
Lemma 2.3. If 1 > > 0 and " > 0, there exists an M( , ") ≥ 1 with the following property. Suppose that n ≥ 2, n ≥ N for sufficiently large n, and X1,X2,...,XN are independent random variables each uniformly distributed on
Γn = {r/n : 1 ≤ r ≤ n}.
Then with probability at least 1 − "/n,
N X Xj ({r/n}) ≤ M( , ") j=1 for all 1 ≤ r ≤ n.
Proof. Since 1 − > 0, there exists an integer M( , ") ≥ 2 such that " n2−M( ,")(1− ) < (2.1) 2 12 for all n ≥ 2. Fix r and set Yj = Xj ({r/n}). Note that Yj are independent random variables, as Xj are. Since Xj are uniformly distributed on Γn, we get
P(Yj = 1) = 1/n,
P(Yj = 0) = 1 − 1/n.
Since 1 > > 0 and n ≥ N, we have n ≥ n ≥ N, yielding Nn−1 ≤ 1. Clearly 0 ≤ Nn−1. Then with p = n−1 and m = M( , ") ≥ 2, Lemma 2.1 tells us that
N ! N ! X X P Xj ({r/n}) ≥ M( , ") = P Yj ≥ M( , ") j=1 j=1 2(Nn−1)M( ,") ≤ M( , ")! 2n M( ,")n−M( ,") ≤ M( , ")! " ≤ 2n−M( ,")(1− ) < , n2 where we used n ≥ N in the second inequality, and the final inequality follows from (2.1). Allowing r to vary, we get immediately
N ! X P Xj ({r/n}) ≥ M( , ") for some 1 ≤ r ≤ n j=1 N N ! X X = P Xj ({r/n}) ≥ M( , ") r=1 j=1 " " < N ≤ , n2 n where we have used the fact that N ≤ n.
Definition 2.4. Suppose P is a probability measure, X is a random variable with respect to the -algebra ℱ, and G ⊆ ℱ is a sub--algebra. If Y is a random variable with respect to the -algebra G such that for all A ∈ G, Z Z X dP = Y dP, A A
13 then we call Y the conditional expectation of X with respect to G, and denote it by E(X ∣ G). It is a consequence of the Radon-Nikodym Theorem that such a random variable always
exists, and is unique almost surely. If X1 is some random variable (with respect to the Borel -algebra ℬ), we can define
E(X ∣ X1) := E(X ∣ (X1)).
−1 Recall (X1) = {X1 (B): B ∈ ℬ} is the smallest -algebra such that X1 is a random
variable. More generally, if X1,...,Xn are random variables, we define
E(X ∣ X1,...,Xn) := E(X ∣ (X1,...,Xn)).
We will require two simple facts about conditional expectation.
Proposition 2.5. (i) If X is G-measurable E(X ∣ G) = X.
(ii) For any X, E(E(X ∣ G)) = E(X).
Proof. Both parts are immediate from the definition.
Lemma 2.6. Let > 0 and for j = 0, 1, . . . , n, let Wj be a random variable that is
measurable with respect to (X1,...,Xj). Define Yj = Wj − Wj−1 and suppose that
2 sYj aj s /2 E(e ∣ X0,X1,...,Xj−1) ≤ e
PN for all ∣s∣ < and some aj ≥ 0. Suppose further that A ≥ j=1 aj. Then provided that 0 ≤ x < A, we have −x2 P(∣W − W ∣ ≥ x) ≤ 2exp . N 0 2A
s(WN −W0) s(WN −WN−1) s(WN−1−W0) Proof. Suppose − < s < . By definition of Wj, we have e = e e .
Since WN−1 and W0 are (X1,...,XN−1)-measurable,
s(WN −W0) s(WN−1−W0) s(WN −WN−1) E(e ∣ X0,X1,...,XN−1) = e E(e ∣ X0,X1,...,XN−1)
s(WN−1−W0) sYN = e E(e ∣ X0,X1,...,XN−1)
14 2 ≤ es(WN−1−W0)eaN s /2.
By property of conditional expectation, taking expectation of both sides gives
2 E(es(WN −W0)) ≤ E(es(WN−1−W0))eaN s /2.
By induction, we get N Y 2 2 E(es(WN −W0)) ≤ eaj s /2 ≤ eAs /2, j=1 PN since A ≥ j=1 aj. Then by Markov’s inequality (see Appendix C.1),
2 s(WN −W0) sx s(WN −W0) −sx As /2 −sx P(WN − W0 ≥ x) = P(e ≥ e ) ≤ E(e )e ≤ e e .
Setting s = xA−1, 0 ≤ x < A ensures that 0 ≤ xA−1 < , so that ∣s∣ < . Then by the above, 2 2 2 −2 −1 x P(W − W ≥ x) ≤ eAs /2e−sx = eAx A /2e−xA x = exp − . N 0 2A
The same argument applies to the sequence −Wj, so that we also have x2 P(W − W ≥ x) ≤ exp − . 0 N 2A The result follows.
Lemma 2.7. Suppose : ℕ → ℝ is a sequence with (n)(log n)1/2 → ∞ as n → ∞, and for any > 0, we have (n)n− → 0 as n → ∞. Suppose 1 > > 1/2, and the positive integer N = N(n) satisfies n ≥ N ≥ n1/2+ for some > 0, for all sufficiently large n.
If " > 0, there exists an M( ) and an n0( , , ") ≥ 1 with the following property. Suppose that n ≥ n0( , , "), and n is odd. Suppose further that X1,X2,...,XN are independent random variables each uniformly distributed on
Γn = {r/n ∈ T : 1 ≤ r ≤ n}.
−1 PN Then, if we write = N j=1 Xj , we have
(n)(log n)1/2 ∣ ∗ ({k/n}) − n−1∣ ≤ " , Nn1/2 15 and M( ) ({k/n}) ≤ N for all 1 ≤ k ≤ n, with probability at least 1/2.
Proof. Set M( ) = M( , 1/4) coming from Lemma 2.3. Without loss of generality, assume Pj−1 M( ) ≥ 3. Fix r for now, and define Y1,Y2,...,YN as follows. If v=1 Xv ({u/n}) ≤ M( ) for all u with 1 ≤ u ≤ n, set
j−1 2j − 1 X Y = − + ({r/n}) + 2 ({r/n}). j n 2Xj Xv+Xj v=1
Pj Otherwise, set Yj = 0. Take X0 = W0 = 0 and Wj = m=1 Ym; notice Yj = Wj − Wj−1. Pj−1 Suppose first that v=1 Xv ({u/n}) ≤ M( ) for all u with 1 ≤ u ≤ n. Observe that if s is a fixed integer, then Xj + s/n and 2Xj are uniformly distributed over Γn (using addition modulo 1), since n is odd and Xj are independent. Then by uniform distribution,
j−1 2j − 1 X E(Y ) = − + E( ({r/n})) + 2 E( ({r/n})) j n 2Xj Xv+Xj v=1 j−1 2j − 1 1 X 1 2j − 2 j − 1 = − + + 2 = − + 2 = 0. n n n n n v=1
Pj−1 On the other hand, if it is not the case that v=1 Xv ({u/n}) ≤ M( ) for all u with
1 ≤ u ≤ n, then Yj = 0 by construction so that E(Yj) = 0 automatically.
sYj We wish to make use of Lemma 2.6, and so we bound E(e ∣ X0,X1,...,Xj−1). Notice as n ≥ N and < 1, we have n > n ≥ N. Then (2N − 1)/n ≤ 2N/n ≤ 2. Pj−1 Suppose first that v=1 Xv ({u/n}) ≤ M( ) for all u with 1 ≤ u ≤ n. Then
j−1 2j − 1 X ∣Y ∣ ≤ + ({r/n}) + 2 ({r/n}) j n 2Xj Xv+Xj v=1 2N − 1 ≤ + 1 + 2M( ) ≤ 2 + 1 + 2M( ) ≤ 3M( ), n
16 where we have used the fact that M( ) ≥ 3. Setting Zj = Yj + (2j − 1)/n, we notice that
Zj ∕= 0 if any of Xv+Xj ({r/n}) ∕= 0 for 1 ≤ v ≤ j. As Xj are uniformly distributed and by subadditivity, we thus get P(Zj ∕= 0) ≤ j/n. Since E(Yj) = 0,
∞ k ! ∞ k X (sYj) X s E(esYj ) = E = E(Y k) k! k! j k=0 k=0 ∞ ∞ X sk X ∣s∣k = 1 + E(Y k) ≤ 1 + E∣Y ∣k k! j k! j k=2 k=2 ∞ ∞ X ∣s∣k X ∣s∣k = 1 + P(Z = 0) E(∣Y ∣k ∣ Z = 0) + P(Z ∕= 0) E(∣Y ∣k ∣ Z ∕= 0) j k! j j j k! j j k=2 k=2 ∞ ∞ X ∣s∣k j X ∣s∣k ≤ 1 + E(∣Y ∣k ∣ Z = 0) + E(∣Y ∣k ∣ Z ∕= 0). k! j j n k! j j k=2 k=2 where the third line follows from the partition formula, and the final inequality follows since P(Zj = 0) ≤ 1 and P(Zj ∕= 0) ≤ j/n. Since Zj = Yj + (2j − 1)/n, in the case Zj = 0 we have ∣Yj∣ = (2j −1)/n. Otherwise, our earlier approximation gives ∣Yj∣ ≤ 3M( ). Thus,
∞ 2j−1 ∞ X (∣s∣ )k j X (∣s∣3M( ))k E(esYj ) ≤ 1 + n + k! n k! k=2 k=2 ∞ ∞ X (∣s∣ 2N−1 )k N X (∣s∣3M( ))k ≤ 1 + n + k! n k! k=2 k=2 since 1 ≤ j ≤ N. Notice if we have a constant 0 ≤ a ≤ 1, then
∞ ∞ X ak X a2 ≤ = a2. k! 2k k=2 k=1
2N−1 1 For sufficiently large n, we have 0 ≤ ∣s∣ n ≤ 1. If ∣s∣ < 3M( ) , we also get 0 ≤ ∣s∣3M( ) ≤ 1, so by our quick result above,
2N − 12 N N 2 N E(esYj ) ≤ 1 + ∣s∣ + (∣s∣3M( ))2 ≤ 1 + 4 s2 + 9M( )2 s2 n n n2 n ≤ 1 + (4 + 9M( )2)Nn−1s2 ≤ 1 + 10M( )2Nn−1s2 ≤ exp 20M( )2Nn−1s2/2 ,
17 where we have used M( )2 ≥ 9 in the second to last inequality, and the fact that 1+x ≤ ex 2 N for every real x in the last inequality. Define aj = 20M( ) n > 0. Pj−1 If it is not the case that v=1 Xv ({u/n}) ≤ M( ) for all u with 1 ≤ u ≤ n, then by
definition of Yj, we get automatically
E(esYj ) = E(e0) = E(1) = 1 ≤ exp 20M( )2Nn−1s2/2 .
Define N N X X 2 −1 2 2 −1 A = aj = 20M( ) Nn = 20M( ) N n . j=1 j=1 1 Set = 3M( ) and x = " (n)(log n)1/2Nn−1/2. To apply Lemma 2.6, we must check that 0 ≤ x < A for sufficiently large n. Indeed, (n)(log n)1/2 → ∞ as n → ∞, while " > 0 and Nn−1/2 > 0 for every n, so that x ≥ 0 for sufficiently large n. In order to show x < A, we must have 1 " (n)(log n)1/2Nn−1/2 < 20M( )2N 2n−1 3M( ) 3 ⇐⇒ " (n)(log n)1/2n1/2 < N. 20M( ) By hypothesis, N is such that there exists > 0 with N > n1/2+ for all sufficiently large −/2 20M( ) n. For this value of > 0, we get (n)n < 3" for sufficiently large n by property of . Then we easily get that for sufficiently large n, 3 " (n)(log n)1/2n1/2 < n/2(log n)1/2n1/2 < n1/2+ < N, 20M( ) as required. Then by Lemma 2.6, x2 "2 (n)2(log n)N 2n−1 P(∣W − W ∣ ≥ x) ≤ 2 exp − = 2 exp − N 0 2A 40M( )2N 2n−1 "2 = 2 exp − (n)2 log n . 40M( )2
"2 2 Notice − 40M( )2 < 0 is a constant, and we are given (n) log n → ∞ as n → ∞. Then,
we may choose n0( , , ") ≥ 1 such that for all n ≥ n0, 1 P(∣W − W ∣ ≥ " (n)(log n)1/2Nn−1/2) ≤ . (2.2) N 0 4 18 For the rest of the proof, we assume n satisfies this condition.
Recall M( ) = M( , 1/4), as given by Lemma 2.3. By that lemma, we have that with probability at least 1 − 1/(4n),
N X Xv ({r/n}) ≤ M( ), (2.3) v=1 for all 1 ≤ r ≤ n. Then for every 1 ≤ r ≤ n and for every 1 ≤ j ≤ N,
j N X X Xv ({r/n}) ≤ Xv ({r/n}) ≤ M( ), v=1 v=1 with probability at least 1 − 1/(4n). Then by construction,
j−1 2j − 1 X Y = − + ({r/n}) + 2 ({r/n}). j n 2Xj Xv+Xj v=1
By definition of Wj,
N j−1 ! X 2j − 1 X W − W = − + ({r/n}) + 2 ({r/n}) . N 0 n 2Xj Xv+Xj j=1 v=1 Notice N X 2j − 1 N 2 − = − , n n j=1 while N j−1 ! N N X X X X 2Xj ({r/n}) + 2 Xv+Xj ({r/n}) = Xv+Xj ({r/n}). (2.4) j=1 v=1 v=1 j=1 Combining (2.2) and (2.4), we get that with probability at least 1 − 1/(4n) − 1/4 ≥ 1/2, we have
N N X X N 2 ∣W − W ∣ = ({r/n}) − < " (n)(log n)1/2Nn−1/2. N 0 Xv+Xj n v=1 j=1
−1 PN Writing = N j=1 Xj , this becomes
(n)(log n)1/2 ∣ ∗ ({r/n}) − n−1∣ < " , Nn1/2 19 and (2.3) becomes M( ) ({r/n}) ≤ . N Allowing r to take values from 1 to n, the result follows.
Lemma 2.8. Suppose , , and N are as in Lemma 2.7. If " > 0, there exists an M( ) and an n0( , , ") ≥ 1 with the following property. Suppose that n ≥ n0( , , "), and n is odd. Then we can find N points
xj ∈ {r/n ∈ T : 1 ≤ r ≤ n}
(not necessarily distinct) such that, writing
N −1 X = N xj , j=1 we have (n)(log n)1/2 ∣ ∗ ({k/n}) − n−1∣ ≤ " , Nn1/2 and M( ) ({k/n}) ≤ N for all 1 ≤ k ≤ n.
Proof. This follows immediately from Lemma 2.7, since an event with positive probability must have at least one occurrence.
20 Chapter 3
A Key Lemma
In this chapter, we will take steps to convert the discrete measure generated by Lemma 2.8 into an incredibly well-behaved function, being a periodic, positive, infinitely differentiable function satisfying several key properties; this procedure will be completed in four lemmas.
We will write 1A for the indicator function of the set A.
Lemma 3.1. Suppose , , and N are as in Lemma 2.7. If " > 0, there exists an M( ) and an n0( , , ") ≥ 1 with the following property. Suppose that n ≥ n0( , , "), and n is odd. Then we can find N points
xj ∈ {r/n ∈ T : 1 ≤ r ≤ n}
(not necessarily distinct) such that, writing
N n X g = 1 −1 −1 , N [xj −(2n) ,xj +(2n) ) j=1 we have:
(n)(log n)1/2n1/2 (i) ∥g ∗ g − 1∥∞ ≤ " N .
(ii) For each t ∈ T, if 0 < ∣ℎ∣ < 1/n, then (n)(log n)1/2n3/2 ∣ℎ∣−1∣g ∗ g(t + ℎ) − g ∗ g(t)∣ ≤ 4" . N
21 nM( ) (iii) ∣g(t)∣ ≤ N for all t ∈ T.
(iv) ∥g∥1 = 1.
Proof. By Lemma2 .8, we can find N points
xj ∈ {r/n ∈ T : 1 ≤ r ≤ n}
(not necessarily distinct) such that, writing
N −1 X = N xj , j=1 we have (n)(log n)1/2 ∣ ∗ ({k/n}) − n−1∣ ≤ " , Nn1/2 and M( ) ({k/n}) ≤ N for all 1 ≤ k ≤ n. Observe that as xj is supported on {xj}, for any a, b and any x ∈ T we have Z
xj ∗ 1[a,b)(x) = 1[a,b)(x − y) dxj (y) = 1[a,b)(x − xj) = 1[xj +a,xj +b)(x). As convolution respects addition,
N N n X n X n ∗ 1 −1 −1 = ∗ 1 −1 −1 = 1 −1 −1 = g. [−(2n) ,(2n) ) N xj [−(2n) ,(2n) ) N [xj −(2n) ,xj +(2n) ) j=1 j=1
Since 1[−(2n)−1,(2n)−1) is symmetric almost everywhere, Z 1[−(2n)−1,(2n)−1) ∗ 1[−(2n)−1,(2n)−1) = 1[−(2n)−1,(2n)−1)(t − x)1[−(2n)−1,(2n)−1)(t) dt Z = 1[x−(2n)−1,x+(2n)−1)(t)1[−(2n)−1,(2n)−1)(t) dt ⎧ 0 if x < −1/n, R ⎨ 1[−(2n)−1,x+(2n)−1)(t) dt if − 1/n ≤ x ≤ 0, = R 1 −1 −1 (t) dt if 0 ≤ x ≤ 1/n, [x−(2n) ,(2n) ) ⎩0 if x > 1/n,
22 ⎧ ⎨0 if ∣x∣ > 1/n, = ⎩1/n − ∣x∣ if ∣x∣ ≤ 1/n, = max {0, 1/n − ∣x∣} .
Writing
Δn = max {0, 1 − n∣x∣} , the previous calculations give
g ∗ g = (n ∗ 1[−(2n)−1,(2n)−1)) ∗ (n ∗ 1[−(2n)−1,(2n)−1)) = ∗ ∗ nΔn.
In particular, we have Z g ∗ g(r/n) = ( ∗ ) ∗ nΔn(r/n) = n max {0, 1 − n∣(r/n) − t∣} d( ∗ )(t).
Since is supported on {k/n}, 1 ≤ k ≤ n, so is ∗ . Then the above integral is zero except possibly when t = r0/n for r0 ∈ ℤ, i.e. nt ∈ ℤ. Now, Δn((r/n) − t) is non-zero only if 1 − n∣(r/n) − t∣ > 0, yielding 1 > ∣r − nt∣. But nt ∈ ℤ, implying nt = r. Therefore,
g ∗ g(r/n) = n(1 − n∣(r/n) − (r/n)∣) ∗ ({r/n}) = n ∗ ({r/n}). (3.1)
It is routine to check that for xj ∈ {r/n ∈ T : 1 ≤ r ≤ n}, we have
−1 −1 −1 −1 1[xi−(2n) ,xi+(2n) ) ∗ 1[xj −(2n) ,xj +(2n) )(x) ⎧ 0 if x∈ / [xi + xj − 1/n, xi + xj + 1/n], ⎨ = x − (xi + xj) + 1/n if x ∈ [xi + xj − 1/n, xi + xj], ⎩−x + (xi + xj) + 1/n if if x ∈ [xi + xj, xi + xj + 1/n].
Notice xi + xj is of the form r/n as well, so that this convolution is continuous and piecewise linear on each interval [(r − 1)/n, r/n], for 1 ≤ r ≤ n. Then g ∗ g, being a linear combination of such convolutions, will be continuous1 and piecewise linear on each
1Continuity can also be obtained by noticing g ∈ L2(T), and L2(T) ∗ L2(T) ⊆ C(T).
23 interval [(r − 1)/n, r/n], for 1 ≤ r ≤ n. Thus, g ∗ g attains a local maximum at points in {r/n ∈ T : 1 ≤ r ≤ n}. By property of the chosen , we have
(n)(log n)1/2 ∣ ∗ ({r/n}) − n−1∣ ≤ " . Nn1/2 By (3.1), this becomes
(n)(log n)1/2n1/2 ∣g ∗ g(r/n) − 1∣ ≤ " . N
Since g ∗ g has local maxima on {r/n ∈ T : 1 ≤ r ≤ n}, we get (n)(log n)1/2n1/2 ∥g ∗ g − 1∥∞ = max ∣g ∗ g(r/n) − 1∣ ≤ " , 1≤r≤n N verifying part (i).
To verify part (ii), we need to check the claimed inequality for each t ∈ T. We have two cases.
Case 1: t = r/n, for some 1 ≤ r ≤ n.
Suppose 0 < ℎ < 1/n. Since g ∗ g is linear on [r/n, (r + 1)/n], g ∗ g(r/n + ℎ) can be expressed as the convex combination:
r r r+1 g ∗ g n + ℎ = (1 − nℎ)(g ∗ g) n + nℎ(g ∗ g) n .
Then
−1 −1 r r ∣ℎ∣ ∣g ∗ g(t + ℎ) − g ∗ g(t)∣ = ∣ℎ∣ ∣g ∗ g( n + ℎ) − g ∗ g( n )∣ −1 r r+1 r = ∣ℎ∣ ∣(1 − nℎ)(g ∗ g)( n ) + nℎ(g ∗ g)( n ) − g ∗ g( n )∣ −1 r+1 r = ∣ℎ∣ n∣ℎ∣∣g ∗ g( n ) − g ∗ g( n )∣ r+1 r ≤ n(∣g ∗ g( n ) − 1∣ + ∣g ∗ g( n ) − 1∣) (n)(log n)1/2n1/2 (n)(log n)1/2n1/2 ≤ n " + " N N (n)(log n)1/2n3/2 = 2" . N
24 The argument is similar if −1/n < ℎ < 0.
Case 2: t ∕= r/n, for any 1 ≤ r ≤ n.
Suppose 0 < ∣ℎ∣ < 1/n. Then there exists r/n, for some 1 ≤ r ≤ n, within a distance of 1/n of both t and t + ℎ. By Case 1,
−1 −1 r r ∣ℎ∣ ∣g ∗ g(t + ℎ) − g ∗ g(t)∣ ≤ ∣ℎ∣ (∣g ∗ g(t + ℎ) − g ∗ g( n )∣ + ∣g ∗ g( n ) − g ∗ g(t)∣) (n)(log n)1/2n3/2 ≤ 4" . N
Cases 1 and 2 together give part (ii).
Fix t ∈ T. We have that [t − (2n)−1, t + (2n)−1) intersects {r/n ∈ T : 1 ≤ r ≤ n} exactly once, say at k/n. Then as is supported on {r/n},
∣g(t)∣ = ∣n ∗ 1[−(2n)−1,(2n)−1)(t)∣ Z = n 1[−(2n)−1,(2n)−1)(t − y) d(y) nM( ) = n({k/n}) ≤ , N by property of , completing part (iii).
By definition of g, it is easy to see that
N Z n X ∥g∥ = 1 −1 −1 (t) dt 1 N [xj −(2n) ,xj +(2n) ) j=1 N n X = ∣(x + (2n)−1) − (x − (2n)−1)∣ N j j j=1 N n X 1 n N = = = 1, N n N n j=1 verifying (iv).
We next smooth the function obtained in the previous lemma:
25 Lemma 3.2. Suppose , , and N are as in Lemma 2.7. If " > 0, there exists an M( )
and an n0( , , ") ≥ 1 with the following property. Suppose that n ≥ n0( , , "), and n is odd. Then we can find a positive, infinitely differentiable function f such that:
(n)(log n)1/2n1/2 (i) ∥f ∗ f − 1∥∞ ≤ " N .
′ (n)(log n)1/2n3/2 (ii) ∥(f ∗ f) ∥∞ ≤ 4" N .
nM( ) (iii) ∥f∥∞ ≤ N .
′ 8n2M( ) (iv) ∥f ∥∞ ≤ N .
(v) R f(t) dt = 1. T (vi) supp f can be covered by N intervals of length 2/n.
Proof. Take M( ), n0( , , "), and g as in Lemma 3.1. Let K : ℝ → ℝ be an infinitely differentiable positive function such that K(x) = 0 for ∣x∣ ≥ 1/2, ∣K′(x)∣ ≤ 8 for all x, and R K(x) dx = 1. (Such a function does exist: as an example, take ℝ k(x) K(x) = , R k(x) dx ℝ where ⎧ 1 1 ⎨exp − 2 if ∣x∣ < , k(x) = 1−4x 2 ⎩0 otherwise. After some work, one can show that K(x) is infinitely differentiable and ∣K′(x)∣ ≤ 7.2 everywhere. It is trivial that it is non-negative and the support lies in (−1/2, 1/2).) Define
Kn : T → ℝ by Kn(t) = nK(nt) for −1/2 ≤ t < 1/2. Then Kn and Kn ∗ Kn are also positive and lie in L1(T), so for each n, we get that
2 Z 1/2 ! 2 ∥Kn ∗ Kn∥1 = ∥Kn∥1 = nK(nt) dt −1/2 2 2 Z n/2 ! Z 1/2 ! = K(x) dx = K(x) dx = 1, −n/2 −1/2
26 since K(x) is supported on (−1/2, 1/2) and n/2 ≥ 1/2 for every n, where equality holds since we have positive functions.
Set f = g ∗ Kn. Then for s ∈ T,
∣(f ∗ f − 1)(s)∣ = ∣(g ∗ Kn) ∗ (g ∗ Kn)(s) − 1∣ Z Z
= (g ∗ g)(s − t)(Kn ∗ Kn)(t) dt − Kn ∗ Kn(t) dt
Z ≤ ∣(g ∗ g)(s − t) − 1∣∣(Kn ∗ Kn)(t)∣ dt
and so
∥f ∗ f − 1∥∞ ≤ ∥g ∗ g − 1∥∞∥Kn ∗ Kn∥1 (n)(log n)1/2n1/2 ≤ " , N verifying part (i).
Fix t ∈ T. By Lemma 3.1, if 0 < ∣ℎ∣ < 1/n, then (n)(log n)1/2n3/2 ∣ℎ∣−1∣g ∗ g(t + ℎ) − g ∗ g(t)∣ ≤ 4" . N Then
∣ℎ∣−1∣f ∗ f(t + ℎ) − f ∗ f(t)∣ −1 = ∣ℎ∣ ∣(g ∗ g) ∗ (Kn ∗ Kn)(t + ℎ) − (g ∗ g) ∗ (Kn ∗ Kn)(t)∣ Z Z −1 = ∣ℎ∣ (g ∗ g)(t + ℎ − y)(Kn ∗ Kn)(y) dy − (g ∗ g)(t − y)(Kn ∗ Kn)(y) dy
Z −1 ≤ ∣ℎ∣ ∣g ∗ g(t + ℎ − y) − g ∗ g(t − y)∣(Kn ∗ Kn)(y) dy Z (n)(log n)1/2n3/2 ≤ 4" (K ∗ K )(y) dy N n n (n)(log n)1/2n3/2 = 4" . N
27 1 Since g, Kn ∈ L and Kn is infinitely differentiable, f = g ∗ Kn is infinitely differentiable. Then f ∗ f is infinitely differentiable, and so the following limit exists: for each t ∈ T, (n)(log n)1/2n3/2 ∣(f ∗ f)′(t)∣ = lim ∣ℎ∣−1∣f ∗ f(t + ℎ) − f ∗ f(t)∣ ≤ 4" , ℎ→0 N and hence (n)(log n)1/2n3/2 ∥(f ∗ f)′∥ ≤ 4" , ∞ N verifying part (ii).
nM( ) Recall ∣g(t)∣ ≤ N for each t ∈ T, so that for s ∈ T, Z Z nM( ) nM( ) ∣f(s)∣ = ∣g ∗ K (s)∣ ≤ ∣g(s − t)∣K (t) dt ≤ K (t) dt = , n n N n N and hence nM( ) ∥f∥ ≤ , ∞ N verifying part (iii).
′ ′ 2 Since Kn is infinitely differentiable, (g ∗ Kn) = g ∗ Kn . By choice of K, for s ∈ T we have Z ′ ′ ′ ′ ∣f (s)∣ = ∣(g ∗ Kn) (s)∣ = ∣g ∗ Kn(s)∣ ≤ ∣g(s − t)∣∣Kn(t)∣ dt Z 1/2 nM( ) Z 1/2 = ∣g(s − t)∣∣n2K′(nt)∣ dt ≤ ∣n2K′(nt)∣ dt −1/2 N −1/2 n2M( ) Z 1/2 n2M( ) Z 1/2 8n2M( ) = ∣K′(x)∣ dx ≤ 8 dx = , N −1/2 N −1/2 N verifying part (iv).
Since g and Kn are positive, Z Z f(t) dt = (g ∗ Kn)(t) dt = ∥g∥1∥Kn∥1 = 1, T verifying part (v).
2 ′ Indeed, we may consider the Fourier transform of each function: for each r ∈ ℤ we have ((g∗Kn) )b(r) = ′ ′ irg\∗ Kn(r) = irgb(r)Kcn(r) = gb(r)Kcn(r) = (g ∗ Kn)b(r).
28 By construction, the support of g can be covered by N intervals of length 1/n. K is supported on (−1/2, 1/2), so that Kn is supported on (−1/(2n), 1/(2n)). Then the support of f = g ∗ Kn is covered the sum of these supports, so is covered by N intervals of length 2/n, for part (vi).
In our proof, we will not need such precise estimates on the bounds, so we modify the previous lemma and introduce a connection with the parameter .
Lemma 3.3. Suppose that 1/2 > −1/2 > > 0. If " > 0, there exists an n1( , , ") ≥ 1 with the following property. Suppose that n ≥ n0( , , "), and n is odd. Then we can find a positive, infinitely differentiable function f such that:
− (i) ∥f ∗ f − 1∥∞ ≤ "n /2.
′ 1− (ii) ∥(f ∗ f) ∥∞ ≤ "n .
(iii) ∥f∥∞ ≤ "n.
′ 2 (iv) ∥f ∥∞ ≤ "n .
(v) R f(t) dt = 1. T (vi) supp f can be covered by less than "n /2 intervals of length 2/n.
(vii) ∣fb(r)∣ ≤ " for all r ∕= 0.
Proof. Without loss of generality, assume " < 1, for it only puts a more severe restriction on the bounds. Set = ( + + 1/2)/2, (n) = log(n), and N = ⌊n⌋. Notice : ℕ → ℝ is a sequence with (n)(log n)1/2 = (log n)3/2 → ∞ as n → ∞ and for any > 0, we have (n)n− = (log n)n− → 0 as n → ∞. We have > > + 1/2, so since 1 > and > 0, we have 1 > > 1/2. N is a function of n taking values in the positive integers. Since = ( + + 1/2)/2, we define = ( − ( + 1/2))/2 > 0 so that = + 1/2 + and = − . For sufficiently large n we get
N(n) = ⌊n⌋ ≥ ⌊n +1/2⌋ ≥ n( /2)+(1/2).
29 For every n, n ≥ ⌊n ⌋ = N. By Lemma 3.2, there exists an M() and an n0( , , ") ≥ 1 with the following property. If n ≥ n0 and n is odd, we can find a positive, infinitely differentiable function f such that:
′ (n)(log n)1/2n1/2 (i) ∥f ∗ f − 1∥∞ ≤ " N .
′ ′ (n)(log n)1/2n3/2 (ii) ∥(f ∗ f) ∥∞ ≤ 4" N .
′ nM() (iii) ∥f∥∞ ≤ N .
′ ′ 8n2M() (iv) ∥f ∥∞ ≤ N .
(v)′ R f(t) dt = 1. T (vi)′ supp f can be covered by N intervals of length 2/n.
This f is the function we want in the lemma. Choose n1( , , ") ≥ n0 sufficiently large so that if n ≥ n1 is odd, then 4(log n)3/2n− < ", (3.2) and 8M()⌊n⌋−1 < ". (3.3)
Using our particular values, 4(log n)3/2n− < " ≤ 1 turns (i)′ into
3/2 1/2 3/2 1/2 3/2 (log n) n (log n) n − (log n) − ∥f ∗ f − 1∥∞ ≤ " ≤ " = "n ≤ "n /2. + 1 + ⌊n ⌋ n 2 n
2 3/2 − We also get ∥f ∗ f − 1∥∞ < " , since 4(log n) n < ". This will be useful in proving part (vii) later.
(ii)′ becomes, after another application of (3.2),
3/2 3/2 3/2 3/2 3/2 ′ (log n) n (log n) n 1− 4(log n) 1− ∥(f ∗ f) ∥∞ ≤ 4" ≤ 4" = "n ≤ "n . + 1 + ⌊n ⌋ n 2 n
Now using (3.3), (iii)′ becomes
nM() ∥f∥ ≤ ≤ "n, ∞ ⌊n⌋
30 and (iv)′ becomes
8n2M() ∥f ′∥ ≤ ≤ "n2. ∞ ⌊n⌋
(v)′ gives what we need immediately.
(vi)′ tells us that supp f can be covered by ⌊n⌋ = ⌊n −⌋ intervals of length 2/n. Our choice of n satisfying (3.2) will ensure that n− < "/2 provided n > exp(1/4) ≈ 1.28, which we will assume. Thus, supp f can be covered by less than n − < "n /2 intervals of length 2/n.
For r ∕= 0,
1/2 1/2 1/2 ∣fb(r)∣ = ∣f[∗ f(r)∣ = ∣f[∗ f(r) − b1(r)∣ ≤ ∥f ∗ f − 1∥∞ < ", verifying part (vii).
For the final step of this chapter, we will impose a periodicity condition on the smooth function we generate.
Lemma 3.4. Suppose that 1/2 > − 1/2 > > 0. There exists an integer k( , ) such that given any " > 0, there exists an m1(k, , , ") ≥ 1 with the following property. Suppose that m > m1(k, , , "), and m is odd. Then we can find a positive, infinitely differentiable, periodic function F with period 1/m with the following properties:
(i) ∥F ∗ F − 1∥∞ ≤ ".
′ k(1− ) (ii) ∥(F ∗ F ) ∥∞ ≤ "m .
k (iii) ∥F ∥∞ ≤ "m .
′ 2k+1 (iv) ∥F ∥∞ ≤ "m .
(v) R F (t) dt = 1. T
31 (vi) We can find a finite collection of intervals ℐ such that [ X I ⊇ supp F and ∣I∣ < ". I∈ℐ I∈ℐ
(vii) ∣Fb(r)∣ ≤ " for all r ∕= 0.
(viii) ∣ℎ∣− ∣F ∗ F (t + ℎ) − F ∗ F (t)∣ ≤ " for all t, ℎ ∈ T with ℎ ∕= 0.
Proof. Since 1/2 > − 1/2 > > 0, we can find 1 and 1 such that 1/2 > − 1/2 >
1 − 1/2 > 1 > > 0. For an explicit choice, let
3 1 1 1 1 3 1 = 4 + 4 + 2 , 1 = 4 − 2 + 4 .
Since 1/2 > − 1/2 > > 0, clearly 1 > 0 and
1 1 1 1 − 2 = 1 + 2 ( − 2 − ) > 1, while 1 1 1 1 1 1 − 2 = ( − 2 ) − 4 ( − 2 − ) < 2 .
Moreover, 1 1 1 = 4 ( − 2 − ) + > , and 1 1 1 = − 4 ( − 2 − ) < .
Choose an integer k such that k( 1 − ) ≥ 1 (as 1 > ) and k( − 1) > 1 − (as
> 1); then k is determined entirely by and alone. By Lemma 3.3, if " > 0 there k exists an n1( 1, 1,") ≥ 1 such that if for some odd m, n = m > n1 (note that n is odd since m is), then we can find a positive, infinitely differentiable function f such that:
′ −k 1 (i) ∥f ∗ f − 1∥∞ ≤ "m /2.
′ ′ k(1− 1) (ii) ∥(f ∗ f) ∥∞ ≤ "m .
′ k (iii) ∥f∥∞ ≤ "m .
32 ′ ′ 2k (iv) ∥f ∥∞ ≤ "m .
(v)′ R f(t) dt = 1. T
(vi)′ supp f can be covered by less than "mk 1 /2 intervals of length 2m−k.
(vii)′ ∣fb(r)∣ ≤ " for all r ∕= 0.
Extend f by periodicity to [−m/2, m/2]. Setting F (t) = f(mt) for t ∈ [−1/2, 1/2], we get that F is a positive, infinitely differentiable, periodic function with period 1/m, since f is a function on T = [−1/2, 1/2] with endpoints identified. We have Z Z 1/2 F ∗ F (s) = F (s − t)F (t) dt = f(m(s − t))f(mt) dt −1/2 Z m/2 1 Z 1/2 = f(ms − u)f(u) du = f(ms − u)f(u) du = f ∗ f(ms), −m/2 m −1/2 by translation invariance of Lebesgue measure. Then F has the following properties:
′′ −k 1 (i) ∥F ∗ F − 1∥∞ ≤ "m /2.
′′ ′ k(1− 1)+1 (ii) ∥(F ∗ F ) ∥∞ ≤ "m .
′′ k (iii) ∥F ∥∞ ≤ "m .
′′ ′ 2k+1 (iv) ∥F ∥∞ ≤ "m .
(v)′′ R F (t) dt = 1. T
(vi)′′ supp F can be covered by less than "mk 1+1/2 intervals of length 2m−k−1.
(vii)′′ ∣Fb(r)∣ ≤ " for all r ∕= 0.
′′ As we chose k such that k( 1 − ) ≥ 1, we get 1 − k 1 ≤ −k , and so (i) gives
−k 1 1−k 1 −k ∥F ∗ F − 1∥∞ ≤ "m /2 ≤ "m /2 ≤ "m /2 ≤ ",
−k verifying (i). We have kept the inequality ∥F ∗ F − 1∥∞ ≤ "m /2 as it is instrumental in proving (viii) later.
33 ′′ Similarly, we have by choice of k that k(1 − ) ≥ k(1 − 1) + 1, and so (ii) gives
′ k(1− 1)+1 k(1− ) ∥(F ∗ F ) ∥∞ ≤ "m ≤ "m ,
verifying (ii). (vi)′′ tells us that there is a finite collection of intervals ℐ (in fact, with at most "mk 1+1/2 elements) such that [ I ⊇ supp F, I∈ℐ and X mk 1+1 ∣I∣ ≤ " (2m−k−1) = "2 −1m1− +k( 1− ). 2 I∈ℐ
Since 1 < and k is fixed satisfying k( − 1) > 1 − , there exists an m1(k, , , ") ≥ 1
−1 1− +k( 1− ) such that if m ≥ m1, then 2 m < 1, so that X ∣I∣ < ", I∈ℐ verifying (vi).
′ k(1− ) We have already proven part (ii), namely that ∥(F ∗ F ) ∥∞ ≤ "m . Then since F ∗ F is smooth, the Mean Value theorem yields
−1 ′ k(1− ) ∣ℎ∣ ∣F ∗ F (t + ℎ) − F ∗ F (t)∣ ≤ ∣(F ∗ F ) ∥∞ ≤ "m .
Thus if ∣ℎ∣ ≤ m−k,
∣ℎ∣− ∣F ∗ F (t + ℎ) − F ∗ F (t)∣ = ∣ℎ∣1− ∣ℎ∣−1∣F ∗ F (t + ℎ) − F ∗ F (t)∣ ≤ "∣ℎ∣1− mk(1− ) ≤ ",
since (1 − ) > 0. If ∣ℎ∣ ≥ m−k, then the inequality we showed in the proof of (i) above gives
− − − −k ∣ℎ∣ ∣F ∗ F (t + ℎ) − F ∗ F (t)∣ ≤ ∣ℎ∣ 2∥F ∗ F − 1∥∞ ≤ "∣ℎ∣ m ≤ ",
since − < 0. This verifies part (viii). The remaining parts (iii), (iv), (v), and (vii) follow directly from their counterparts (iii)′′, (iv)′′, (v)′′, and (vii)′′.
34 Chapter 4
The Main Theorem
Our method of proof will, in fact, result in a slightly more general version than what we need for Theorem 4.19. For the sake of generality, we will introduce the concept of - Lipschitz functions in the following section, and develop a complete metric space to work in so that we may eventually apply a Baire Category argument.
4.1 -Lipschitz functions
Lemma 4.1. (i) Consider the space ℱ of non-empty closed subsets of T. If we set dℱ (E,F ) = max sup inf ∣e − f∣, sup inf ∣e − f∣ , e∈E f∈F f∈F e∈E
then (ℱ, dℱ ) is a complete metric space. dℱ is known as the Hausdorff metric.
(ii) Consider the space ℰ consisting of ordered pairs (E, ) where E ∈ ℱ and is a probability measure with supp ⊆ E and b(r) → 0 as ∣r∣ → ∞. If we take
dℰ ((E, ), (F, )) = dℱ (E,F ) + sup ∣b(r) − b(r)∣, r∈ℤ
then (ℰ, dℰ ) is a complete metric space.
35 (iii) Consider the space G consisting of those (E, ) ∈ ℰ such that ∗ = f with f continuous. If we take
dG((E, ), (F, )) = dℰ ((E, ), (F, )) + ∥f − f∥∞,
then (G, dG) is a complete metric space.
Proof. (i) It is a standard result that the Hausdorff metric defined with usual distance ∣ ⋅ ∣ is a complete metric (see Appendix D.3).
(ii) If (E, ), (F, ) ∈ ℰ with E = F and = , then it is clear that dℰ ((E, ), (F, )) = 0. Conversely, if d ((E, ), (F, )) = d (E,F ) + sup ∣(r) − (r)∣ = 0, then E = F by ℰ ℱ r∈ℤ b b (i) and = by the uniqueness theorem.
Since dℱ is a metric, dℰ ((E, ), (F, )) = dℰ ((F, ), (E, )) for any (E, ), (F, ) ∈ ℰ. Triangle inequality follows directly from the usual triangle inequality on ∣ ⋅ ∣ and part
(i). Thus, (ℰ, dℰ ) is indeed a metric space.
To see completeness, let (En, n) be a Cauchy sequence in (ℰ, dℰ ). Then {En} is
Cauchy in (ℱ, dℱ ), so by completeness, we can find a set E ∈ ℱ such that dℱ (En,E) → 0 as n → ∞. Since the space of probability measures is weak-∗ compact, there exists a convergent subsequence, say nk → w-∗, a probability measure. Since T is compact, this implies (r) → (r) for every fixed r. By hypothesis, (r) → 0 as ∣r∣ → ∞, and cnk b cnk by the Cauchy condition, sup ∣ (r) − (r)∣ → 0 cnk cnl r∈ℤ as k, l → ∞, and thus b(r) → 0 as ∣r∣ → ∞.
Suppose x ∈ supp . By definition, for every neighborhood Ux of x, we have (Ux) > 0.
Since n → w-∗, there exists an integer Nx sufficiently large that n ≥ Nx implies
n(U) ≥ (U) for every open set U. In particular, for every neighborhood Ux of x, this gives n(Ux) ≥ (Ux) > 0, so that x ∈ supp n if n ≥ Nx. Let " > 0. By property of Hausdorff metric, (see Appendix D.4), we have that there exists N" such that n ≥ N" implies En ⊆ E". Combined with the previous paragraph, this gives x ∈ supp n ⊆ En ⊆ E" for n ≥ max{Nx,N"}. Since this holds for arbitrary " and E is closed, we have that x ∈ E, and so supp ⊆ E.
36 Weak convergence gives (r) → (r) for each fixed r, thus cnk b
dℰ ((Enk , nk ), (E, )) → 0 as k → ∞. Since a Cauchy sequence with a convergent subsequence must converge to the same limit as the subsequence,
dℰ ((En, n), (E, )) → 0, completing the proof.
(iii) That (G, dG) is a metric follows a similar proof as in (ii).
To see completeness, let (En, n) be a Cauchy sequence in (G, dG). Then (En, n) is
Cauchy in (ℰ, dℰ ), so by completeness, we can find (E, ) ∈ ℰ such that dℰ ((En, n), (E, )) →
0 as n → ∞. Since fn is Cauchy in the uniform norm, fn converges uniformly to some continuous function f. By properties of weak convergence, ∗ = f, so (E, ) ∈ G and dG((En, n), (E, )) → 0.
Definition 4.2. Let 1 > > 1/2 and suppose that : ℝ+ → ℝ+ is a strictly increasing, continuous function with (t) ≥ t −1/2 for all t ≥ 0, (0) = 0. We call a generalized -power function. For our uses, need only be defined on [0, 1/2] since we are on the torus and will be using non-negative values.
Definition 4.3. Suppose is a generalized -power function. We say that f : T → ℂ is -Lipschitz if sup (∣ℎ∣)−1∣f(t + ℎ) − f(t)∣ < ∞. t,ℎ∈T,ℎ∕=0
In this case, we denote this value by ! (f); notice ! (f) ≥ 0. In the particular case of (t) = t , this definition reduces to the usual definition of Lipschitz of class , recall (1.2).
We denote by Λ the collection of all -Lipschitz functions. Analogously to Lipschitz functions (see Lemma 1.4), we have a result linking this definition to another form which may be more useable in certain circumstances, with the same proof.
Lemma 4.4. f ∈ Λ if and only if ∣f(x) − f(y)∣ ≤ C (∣x − y∣) for a constant C, for every x, y ∈ T.
37 We develop some basic properties of operations under ! which will be used later.
Lemma 4.5. Let be a generalized -power function. Suppose f, g ∈ Λ , c is a constant, and F ∈ L1(T). Then the following hold:
(i)Λ -translation invariance: c + f ∈ Λ and ! (f + c) = ! (f).
(ii)Λ -positive homogeneity: cf ∈ Λ and ! (cf) = ∣c∣! (f).
(iii)Λ -addition: f + g ∈ Λ and ! (f + g) ≤ ! (f) + ! (g).
(iv)Λ -multiplication: fg ∈ Λ and ! (fg) ≤ ! (f)∥g∥∞ + ! (g)∥f∥∞.
(v)Λ -convolution: f ∗ F ∈ Λ and ! (f ∗ F ) ≤ ! (f)∥F ∥1.
Proof. (i) and (ii) are immediate from the definition.
(iii) This follows by triangle inequality.
(iv) If f, g ∈ Λ , we have
sup (∣ℎ∣)−1∣(fg)(t + ℎ) − (fg)(t)∣ t,ℎ∈T,ℎ∕=0 ≤ sup (∣ℎ∣)−1(∣f(t + ℎ) − f(t)∣∣g(t + ℎ)∣ + ∣f(t)∣∣g(t + ℎ) − g(t)∣) t,ℎ∈T,ℎ∕=0 ≤ sup (∣ℎ∣)−1∣f(t + ℎ) − f(t)∣∣g(t + ℎ)∣ + sup (∣ℎ∣)−1∣f(t)∣∣g(t + ℎ) − g(t)∣ t,ℎ∈T,ℎ∕=0 t,ℎ∈T,ℎ∕=0 −1 −1 ≤ sup (∣ℎ∣) ∣f(t + ℎ) − f(t)∣∥g∥∞ + sup (∣ℎ∣) ∥f∥∞∣g(t + ℎ) − g(t)∣ t,ℎ∈T,ℎ∕=0 t,ℎ∈T,ℎ∕=0
= ! (f)∥g∥∞ + ! (g)∥f∥∞ < ∞, so that fg ∈ Λ and
! (fg) ≤ ! (f)∥g∥∞ + ! (g)∥f∥∞.
1 (v) If f ∈ Λ and F ∈ L (T), we have for fixed t, ℎ ∈ T, ℎ ∕= 0:
(∣ℎ∣)−1∣(f ∗ F )(t + ℎ) − (f ∗ F )(t)∣ Z Z −1 = (∣ℎ∣) f(t + ℎ − y)F (y) dy − f(t − y)F (y) dy
38 Z ≤ (∣ℎ∣)−1 ∣f(t + ℎ − y) − f(t − y)∣∣F (y)∣ dy Z ≤ ! (f)∣F (y)∣ dy = ! (f)∥F ∥1 < ∞, so that taking the supremum over permissible t and ℎ yields f ∗ F ∈ Λ and
! (f ∗ F ) ≤ ! (f)∥F ∥1.
There is also a nice relationship between the ! value of a function and the magnitude of its derivative, provided f is continuously differentiable.
Lemma 4.6. Let be a generalized -power function. If f : T → ℂ has a continuous ′ derivative, then f ∈ Λ and ! (f) ≤ ∥f ∥∞.
′ ′ Proof. If f : T → ℂ has continuous derivative, then ∣f ∣ has a maximum on T, so ∥f ∥∞ < ∞. By the Mean Value theorem, for any t, ℎ ∈ T, ℎ ∕= 0, we have −1 ′ ∣ℎ∣ ∣f(t + ℎ) − f(t)∣ ≤ ∥f ∥∞, so then ∣ℎ∣ ≤ (∣ℎ∣) yields
−1 −1 ′ (∣ℎ∣) ∣f(t + ℎ) − f(t)∣ ≤ ∣ℎ∣ ∣f(t + ℎ) − f(t)∣ ≤ ∥f ∥∞.
Taking the supremum over t, ℎ ∈ T with ℎ ∕= 0, we get −1 ′ sup (∣ℎ∣) ∣f(t + ℎ) − f(t)∣ ≤ ∥f ∥∞ < ∞ t,ℎ∈T,ℎ∕=0 so that f ∈ Λ and ′ ! (f) ≤ ∥f ∥∞.
Lemma 4.7. Let 1 > > 1/2 and suppose that is a generalized -power function.
Consider the space ℒ consisting of those (E, ) ∈ G such that ∗ = f with f ∈ Λ . If we take
d ((E, ), (F, )) = dG((E, ), (F, )) + ! (f − f), then (ℒ , d ) is a complete metric space.
39 Proof. If (E, ), (F, ) ∈ ℒ with E = F and = , then it is clear that d ((E, ), (F, )) =
0. Conversely, if d ((E, ), (F, )) = dG((E, ), (F, )) + ! (f − f) = 0, then E = F and
= since dG is a metric.
Since dG is a metric and by Λ -positive homogeneity, d ((E, ), (F, )) = d ((F, ), (E, )) for any (E, ), (F, ) ∈ ℒ .
Triangle inequality follows directly from Λ -addition and the fact that dG is a metric.
Thus, (ℒ , d ) is indeed a metric space.
If (En, n) is a Cauchy sequence in (ℒ , d ), then it is Cauchy in (G, dG). Thus there exists (E, ) ∈ G such that
dG((En, n), (E, )) → 0 as n → ∞. By definition of G, we may write n ∗ n = fn and ∗ = f with fn and f continuous. The condition with dG above gives fn → f uniformly. If m ≥ n, we have for any t, ℎ ∈ T, ℎ ∕= 0 that
−1 (∣ℎ∣) ∣(f − fn)(t + ℎ) − (f − fn)(t)∣ −1 −1 ≤ (∣ℎ∣) ∣(f − fm)(t + ℎ) − (f − fm)(t)∣ + (∣ℎ∣) ∣(fm − fn)(t + ℎ) − (fm − fn)(t)∣ −1 ≤ (∣ℎ∣) ∣(f − fm)(t + ℎ) − (f − fm)(t)∣ + d ((En, n), (Em, m)) −1 ≤ (∣ℎ∣) ∣(f − fm)(t + ℎ) − (f − fm)(t)∣ + sup d ((Ep, p), (Eq, q)). p,q≥n
Allowing m → ∞, the first term limits to 0 and thus
−1 (∣ℎ∣) ∣(f − fn)(t + ℎ) − (f − fn)(t)∣ ≤ sup d ((Ep, p), (Eq, q)) p,q≥n for all t, ℎ ∈ T, ℎ ∕= 0. Thus f − fn ∈ Λ and so Λ -addition gives f ∈ Λ , implying
(E, ) ∈ ℒ . Moreover, the above calculation gives
! (f − fn) ≤ sup d ((Ep, p), (Eq, q)) p,q≥n so that
d ((En, n), (E, )) → 0 as n → ∞, and hence (ℒ , d ) is a complete metric space.
40 Lemma 4.8. Let 1 > > 1/2 and suppose that is a generalized -power function.
Consider the set {(E, ) ∈ ℒ : ∗ = f, f is infinitely differentiable}. Let ℳ denote the closure of this set with respect to the d metric. Then (ℳ , d ) is a complete metric space.
Proof. By definition, (ℳ , d ) is a closed subspace of the complete metric space (ℒ , d ) (by Lemma 4.7), and hence is complete.
4.2 Density results
Let 1 > > 1/2 and suppose that is a generalized -power function. Let ℋn be the subset of (ℳ , d ) consisting of those (E, ) ∈ ℳ such that we can find a finite collection of closed intervals ℐ with
[ X 1 I ⊇ E and ∣I∣ +1/n < . n I∈ℐ I∈ℐ
Notice that ℋn ⊇ ℋn+1. Our goal will be to show that ℋn are dense in (ℳ , d ) which will be Lemma 4.16.
Lemma 4.9. Let 1 > > 1/2 and suppose that is a generalized -power function. Suppose further that n ≥ 1, g : T → ℝ is a positive, infinitely differentiable function with Z g(t) dt = 1 T and H is a closed set with H ⊇ supp g. Then, given " > 0, we can find a positive, infinitely differentiable function f : T → ℝ with Z f(t) dt = 1 T and a closed set E ⊇ supp f such that (E, f) ∈ ℋn and
d ((E, f), (H, g)) < ".
41 Proof. Since ℋn ⊇ ℋn+1 and 1 > > 1/2, we may restrict ourselves to the case when 1 > + 1/n > 1/2. Lemma 3.4 provides us with an integer k = k( + 1/n, − 1/2) ≥ 1 with the property described in the next sentence. Fix 0 < < 1/n for the time being; then
there exists an integer m1(k, + 1/n, − 1/2, ) ≥ 1 such that if 2m + 1 > m1, we can find
a positive, infinitely differentiable function Fm which is periodic with period 1/(2m + 1) with the following properties:
(i)m ∥Fm ∗ Fm − 1∥∞ ≤ .
′ k(1− +1/2) (ii)m ∥(Fm ∗ Fm) ∥∞ ≤ (2m + 1) .
k k k k (iii)m ∥Fm∥∞ ≤ (2m + 1) ≤ (2(2m)) = 4 m .
′ 2k+1 2k+1 2k+1 2k+1 (iv)m ∥Fm∥∞ ≤ (2m + 1) ≤ (2(2m)) = 4 m . R (v)m ∥Fm∥1 = Fm(t) dt = 1. T
(vi)m We can find a finite collection of intervals ℐm such that
[ X + 1 1 I ⊇ supp F and ∣I∣ n < < . m n I∈ℐm I∈ℐm
(vii)m ∣Fcm(r)∣ ≤ for all r ∕= 0.
− +1/2 (viii)m ∣ℎ∣ ∣Fm ∗ Fm(t + ℎ) − Fm ∗ Fm(t)∣ ≤ for all t, ℎ ∈ T with ℎ ∕= 0.
Since is a generalized -power function, (t) ≥ t −1/2 for every t ≥ 0. Then using
(viii)m and taking the supremum over t, ℎ ∈ T with ℎ ∕= 0,
−1 ! (Fm ∗ Fm) = sup (∣ℎ∣) ∣Fm ∗ Fm(t + ℎ) − Fm ∗ Fm(t)∣ t,ℎ∈T,ℎ∕=0 − +1/2 ≤ sup ∣ℎ∣ ∣Fm ∗ Fm(t + ℎ) − Fm ∗ Fm(t)∣ ≤ . t,ℎ∈T,ℎ∕=0
We shall also call this property (viii)m, as we will no longer need the previous version.
42 Since g is infinitely differentiable, all of its derivatives g(j) are continuous on the torus. In particular, g(j)(−1/2) = g(j)(1/2) for every j. Then by integration by parts with u = g(j)(t) and dv = e−irt dt, we have
Z 1/2 1 (j) (j) −2irt [(j+1) gc (r) = g (t)e dt = g (r). −1/2 ir Then by induction, for every j we have 1 g(r) = gc(j)(r). b (ir)j
In particular, for j = 2k + 4, we have that for r ∕= 0 Z 1 (j) −2irt −(2k+4) ∣g(r)∣ = g e dt ≤ C1∣r∣ , b ∣r∣2k+4
(j) where C1 = ∥g ∥1 is a constant. By a straightforward application of the integral test,
X X 1 ∣r∣∣g(r)∣ ≤ C ∣r∣−(2k+3) ≤ 2C 1 + m−(2k+2) , b 1 1 2k + 2 ∣r∣≥m ∣r∣≥m
2k+3 since m ≥ 1. Thus, we get that with C = 2C1 k+1 a constant,
X −(2k+2) ∣r∣∣gb(r)∣ ≤ Cm . (4.1) ∣r∣≥m
Since g(0) = R g(t) dt = 1, we have in particular that b T ∞ X X ∣gb(r)∣ ≤ ∣gb(0)∣ + ∣r∣∣gb(r)∣ ≤ 1 + C. (4.2) r=−∞ ∣r∣≥1
Set Gm(t) = g(t)Fm(t). Since Fm is periodic of period 1/(2m + 1), we know that R Fm(r) = 0 if r∈ / (2m + 1) . Then since g(0) = 1 and Fm(0) = Fm(t) dt = 1, c ℤ b c T ∞ X Gcm(0) − 1 = gb ∗ Fcm(0) − 1 = gb(r)Fcm(−r) − 1 r=−∞ ∞ X = gb(0)Fcm(0) − 1 + gb((2m + 1)j)Fcm(−(2m + 1)j) j=−∞,j∕=0
43 X = gb(r)Fcm(−r). ∣r∣≥2m+1
Also, ∣Fcm(−r)∣ ≤ ∥Fm∥1 = 1, so that the previous bound of Gcm(0) − 1 gives
X −(2k+2) −(2k+2) ∣Gcm(0) − 1∣ ≤ ∣r∣∣gb(r)∣ ≤ C(2m + 1) ≤ Cm , (4.3) ∣r∣≥2m+1 R R using (4.1). Since Gm(0) = Gm(t) dt,(4.3) gives that for sufficiently large m, Gm(t) dt ∕= c T T 0, so that we may define Z −1 f(t) = Gm(s) ds Gm(t). T This function f will be the function we want to approximate the given g, but we will need in addition a special closed set E that is close to H.
Notice
supp f ⊆ supp g ∩ supp Fm ⊆ H ∩ supp Fm ⊆ H. In particular, sup inf ∣e − ℎ∣ = 0. e∈supp f ℎ∈H H ⊆ T is a closed subset of a compact set, hence compact. Then as (T, ∣ ⋅ ∣) is a metric space, H is totally bounded. That is, (with " > 0 given in the hypothesis) there exist
finitely many balls of radius ri < "/4 centered at points ai, say B1(a1, r1),...,B(aN , rN ), SN SN such that H ⊆ i=1 B(ai, ri). Consider the finite set A = i=1{ai}, and fix ℎ0 ∈ H. Then SN ℎ0 ∈ i=1 B(ai, ri), so that there is some 1 ≤ i ≤ N with ℎ0 ∈ B(ai, ri). In particular, we get an element ai ∈ A with ∣ℎ0 − ai∣ < "/4. Then " inf ∣ℎ0 − a∣ ≤ ∣ℎ0 − ai∣ < . a∈A 4 Taking the supremum over elements in H, " sup inf ∣ℎ − a∣ ≤ . ℎ∈H a∈A 4 Define E := A∪supp f. Then the previous estimate remains unchanged when A is replaced
by E, since for each fixed element ℎ0 in H, the infimum of distances to an element in E
is still smaller than the distance between ℎ0 and ai. E is closed, being the union of
44 finitely many closed sets, so E ∈ ℱ. f is a probability measure since by construction, R f(t) dt = 1, and clearly supp f ⊆ E. f(r) → 0 as ∣r∣ → ∞ since f is infinitely T b differentiable, and so (E, f) ∈ ℰ. f ∗ f = (f ∗ f)( ∗ ) = (f ∗ f), and f ∗ f is continuous (in fact, infinitely differentiable) as f is, so that (E, f) ∈ G. f ∈ Λ as f is infinitely differentiable, and so (E, f) ∈ ℳ . The finite collection of intervals ℐ consisting of those in ℐm together with the degenerate closed intervals [ai, ai] = {ai} has the property that [ X 1 I ⊇ A ∪ supp F ⊇ A ∪ supp f and ∣I∣ +1/n < < , m n I∈ℐ I∈ℐ so that (E, f) ∈ ℋn. Moreover, " " dℱ (E,H) = sup inf ∣e − ℎ∣ + sup inf ∣e − ℎ∣ ≤ 0 + = . (4.4) e∈E ℎ∈H ℎ∈H e∈E 4 4
Recall we are to show that
d ((E, f), (H, g)) < ".
By definition, the metric d is expanded as
d ((E, f), (H, g))
= dG((E, f), (H, g)) + ! (f ∗ f − g ∗ g)
= dℰ ((E, f), (H, g)) + ∥f ∗ f − g ∗ g∥∞ + ! (f ∗ f − g ∗ g)
= dℱ (E,H) + sup ∣fb(r) − gb(r)∣ + ∥f ∗ f − g ∗ g∥∞ + ! (f ∗ f − g ∗ g). r∈ℤ We have bounded the first term by "/4, it remains to show the remaining three terms are also bounded by "/4. In order to show these, we first will show that f is close to Gm for sufficiently large m, and appeal to a triangle inequality argument.
−(2k+2) For ease of notation, write Cm = Cm > 0. We may assume Cm < 1/2 for all sufficiently large m. With this notation, (4.3) yields
1 − Cm ≤ Gcm(0) ≤ 1 + Cm ≤ 2.
Then
−2 ∣1 + Gcm(0)∣∣1 − Gcm(0)∣ 3Cm 3Cm ∣Gcm(0) − 1∣ = ≤ ≤ = 12Cm, 2 2 2 ∣Gcm(0) ∣ (1 − Cm) (1 − 1/2)
45 since Cm < 1/2. As an immediate consequence, we have
−1 −2 −(2k+2) ∣Gcm(0) − 1∣ ≤ ∣Gcm(0) − 1∣ ≤ 12Cm , (4.5)
since ∣x−1 − 1∣ ≤ ∣x−2 − 1∣ for x > 0.
Claim 4.10. " sup ∣fb(r) − Gcm(r)∣ ≤ . r∈ℤ 8
Proof. Fix r ∈ ℤ. Notice that as Gm is the product of two positive functions, it is itself R positive and hence ∥Gm∥1 = Gm(t) dt = Gm(0) ≤ 2. As ∣Gm(r)∣ ≤ ∥Gm∥1, by (4.5) we T c c get
Z −1
∣fb(r) − Gcm(r)∣ = Gm(s) ds Gcm(r) − Gcm(r) T
Z −1 −(2k+2) = ∣Gcm(r)∣ Gm(s) ds − 1 ≤ 24Cm . T Since 24C is a constant, m can be taken sufficiently large that " ∣fb(r) − Gcm(r)∣ < . 8
Taking the supremum over r ∈ ℤ yields the desired result.
Claim 4.11. " ∥f ∗ f − G ∗ G ∥ < . m m ∞ 8
Proof. A direct computation gives, using (4.5),
Z −2
∥f ∗ f − Gm ∗ Gm∥∞ = Gm(s) ds Gm ∗ Gm − Gm ∗ Gm T ∞
Z −2
≤ ∥Gm∥1∥Gm∥∞ Gm(s) ds − 1 T −(2k+2) ≤ 2∥Gm∥∞12Cm .
46 We have ∥Gm∥∞ = ∥gFm∥∞ ≤ ∥g∥∞∥Fm∥∞, and our estimate from (iii)m gives
k k −(2k+2) k −(k+2) ∥f ∗ f − Gm ∗ Gm∥∞ ≤ 24C∥g∥∞4 m m = 24C∥g∥∞4 m .
k Since 24C∥g∥∞4 is a constant, m can be taken sufficiently large that " ∥f ∗ f − G ∗ G ∥ < . m m ∞ 8
Claim 4.12. " ! (f ∗ f − G ∗ G ) < . m m 8 Proof. We have
! (Gm ∗ Gm) ≤ ! (Gm)∥Gm∥1 by Λ -convolution
= ! (gFm)∥Gm∥1 ≤ 2! (gFm)
≤ 2 [! (g)∥Fm∥∞ + ! (Fm)∥g∥∞] by Λ -multiplication ′ ′ ≤ 2 [∥g ∥∞∥Fm∥∞ + ∥Fm∥∞∥g∥∞] by Lemma4 .6 ′ k k 2k+1 2k+1 ≤ 2 ∥g ∥∞4 m + 4 m ∥g∥∞ by (iii)m and (iv)m ′ 2k+1 2k+1 ≤ 2(∥g ∥∞ + ∥g∥∞)4 m .
Then using Λ -positive homogeneity and (4.5), Z −2 ! ! (f ∗ f − Gm ∗ Gm) = ! Gm(s) ds Gm ∗ Gm − Gm ∗ Gm T
Z −2
= Gm(s) ds − 1 ! (Gm ∗ Gm) T −(2k+2) ′ 2k+1 2k+1 ≤ 12Cm 2(∥g ∥∞ + ∥g∥∞)4 m ′ 2k+1 −1 = 24C(∥g ∥∞ + ∥g∥∞)4 m .
′ 2k+1 Since 24C(∥g ∥∞ + ∥g∥∞)4 is a constant, m can be taken sufficiently large that " ! (f ∗ f − G ∗ G ) < . m m 8
47 We next need to show that Gm is close to g for sufficiently large m.
Claim 4.13. " sup ∣gb(r) − Gcm(r)∣ ≤ . r∈ℤ 8
Proof. For a fixed r ∈ ℤ, we have
∞ X ∣gb(r) − Gcm(r)∣ = gb(r) − gb(r − j)Fcm(j) j=−∞
X = gb(r − j)Fcm(j) since Fcm(0) = 1 j∕=0 X ≤ ∣gb(r − j)∣ by (vii)m j∕=0 ≤ (1 + C) by (4.2).
We may choose < "/(8(1 + C)), and so taking supremum over all r ∈ ℤ gives " sup ∣gb(r) − Gcm(r)∣ ≤ . r∈ℤ 8
We now fix for the remainder of the proof, so that the previous claim holds and we have " < (∥g∥ + ! (g ∗ g) + 2)−2 , (4.6) ∞ 24 but we leave m free for now, subject only to the constraint that previous claims remain true.
Claim 4.14. " ∥g ∗ g − G ∗ G ∥ < . m m ∞ 8
48 Proof. Consider the mth partial Fourier sum,
X 2irt Sm(t) = gb(r)e . ∣r∣ Since g is infinitely differentiable on T, the Fourier inversion theorem tells us that ∞ X 2irt g(t) = gb(r)e , r=−∞ and so X 2irt X X ∣(g − Sm)(t)∣ = g(r)e ≤ ∣g(r)∣ ≤ ∣r∣∣g(r)∣. b b b ∣r∣≥m ∣r∣≥m ∣r∣≥m Then by (4.1), −(2k+2) ∥g − Sm∥∞ ≤ Cm , (4.7) and hence we may choose m sufficiently large that ∥g − Sm∥∞ ≤ 1. (4.8) Similarly, ′ X 2irt X ∣(g − Sm) (t)∣ = g(r)(2ir)e ≤ 2 ∣r∣∣g(r)∣, b b ∣r∣≥m ∣r∣≥m so by (4.1), ′ −(2k+2) ∥(g − Sm) ∥∞ ≤ 2Cm . (4.9) Consider the Fourier coefficients of SmFm. For u, v ∈ ℤ with 0 ≤ v ≤ 2m, ∞ X S\mFm((2m + 1)u + v) = Scm(j)Fcm((2m + 1)u + v − j). j=−∞ Since Fm is periodic of period 1/(2m+1), Fcm((2m+1)u+v−j) = 0 unless (2m+1)u+v−j ∈ (2m + 1)ℤ, that is, unless v − j ∈ (2m + 1)ℤ. On the other hand, Sm is a trigonometric polynomial of degree at most m, and hence Scm(j) = 0 unless ∣j∣ < m. Since 0 ≤ v ≤ 2m, we only get non-zero terms if j = v. Thus we have S\mFm((2m + 1)u + v) = Scm(v)Fcm((2m + 1)u). 49 Likewise, we can consider the Fourier coefficients of (Sm ∗Sm)(Fm ∗Fm). Since Sm ∗Sm is a trigonometric polynomial of degree at most m and Fm ∗Fm is periodic of period 1/(2m+1), the above work shows 2 2 ((Sm ∗ Sm)(Fm ∗ Fm))b((2m + 1)u + v) = (Scm(v)) (Fcm((2m + 1)u)) . These last two identities give 2 ((SmFm) ∗ (SmFm))b((2m + 1)u + v) = (S\mFm((2m + 1)u + v)) 2 2 = (Scm(v)) (Fcm((2m + 1)u)) = ((Sm ∗ Sm)(Fm ∗ Fm))b((2m + 1)u + v). Since u, v ∈ ℤ and 0 ≤ v ≤ 2m, every Fourier coefficient of these functions agree, so the uniqueness theorem tells us that (SmFm) ∗ (SmFm)(t) = (Sm ∗ Sm)(t)(Fm ∗ Fm)(t), (4.10) for every t ∈ T. Then ∥g ∗ g − Gm ∗ Gm∥∞ = ∥g ∗ g − (gFm) ∗ (gFm)∥∞ ≤ ∥g ∗ g − Sm ∗ Sm∥∞ + ∥Sm ∗ Sm − (SmFm) ∗ (SmFm)∥∞ + ∥(SmFm) ∗ (SmFm) − (gFm) ∗ (gFm)∥∞ = ∥g ∗ g − Sm ∗ Sm∥∞ + ∥Sm ∗ Sm − (Sm ∗ Sm)(Fm ∗ Fm)∥∞ + ∥(SmFm) ∗ (SmFm) − (gFm) ∗ (gFm)∥∞. Let us consider each of these three terms separately. We rewrite the first term, ∥g ∗ g − Sm ∗ Sm∥∞ = ∥(g + Sm) ∗ (g − Sm)∥∞ ≤ ∥g + Sm∥1∥g − Sm∥∞ ≤ (∥g∥1 + ∥Sm∥1)∥g − Sm∥∞ ≤ (1 + ∥Sm∥∞)∥g − Sm∥∞ −(2k+2) ≤ (2 + ∥g∥∞)(Cm ), where in the last inequality we used (4.7) and (4.8). Since (2 + ∥g∥∞)C is a constant, we can take sufficiently large m that " ∥g ∗ g − S ∗ S ∥ < . (4.11) m m ∞ 24 50 For the second term, ∥Sm ∗ Sm − (Sm ∗ Sm)(Fm ∗ Fm)∥∞ = ∥(Sm ∗ Sm)(1 − Fm ∗ Fm)∥∞ ≤ ∥Sm ∗ Sm∥∞∥1 − Fm ∗ Fm∥∞ 2 ≤ ∥Sm∥∞∥1 − Fm ∗ Fm∥∞ 2 ≤ (1 + ∥g∥∞) , by (4.8) and (i)m. Recall that was chosen so that (4.6) holds, that is, " " < (∥g∥ + ! (g ∗ g) + 2)−2 ≤ (∥g∥ + 1)−2 , ∞ 24 ∞ 24 since ! (g ∗ g) ≥ 0. Thus, " ∥S ∗ S − (S ∗ S )(F ∗ F )∥ < . (4.12) m m m m m m ∞ 24 For the third term, we proceed as in the first case, ∥(SmFm) ∗ (SmFm) − (gFm) ∗ (gFm)∥∞ ≤ ∥SmFm + gFm∥1∥SmFm − gFm∥∞ ≤ ∥Sm + g∥∞∥Fm∥1∥Sm − g∥∞∥Fm∥∞ −(2k+2) k k ≤ (1 + 2∥g∥∞)Cm (4 m ) k −(k+2) = 4 (1 + 2∥g∥∞)Cm , k where we have used (iii)m, (v)m,(4.7), and (4.8) in the last inequality. Since 4 (1+2∥g∥∞)C is a constant, we can take sufficiently large m that " ∥(S F ) ∗ (S F ) − (gF ) ∗ (gF )∥ < . (4.13) m m m m m m ∞ 24 Equations (4.11), (4.12) and (4.13) together give " ∥g ∗ g − G ∗ G ∥ < . m m ∞ 8 Claim 4.15. " ! (g ∗ g − G ∗ G ) < . m m 8 51 Proof. This claim proceeds much like Claim 4.14. By Λ -addition, ! satisfies the triangle inequality: ! (g ∗ g − Gm ∗ Gm) = ! (g ∗ g − (gFm) ∗ (gFm)) ≤ ! (g ∗ g − Sm ∗ Sm) + ! (Sm ∗ Sm − (SmFm) ∗ (SmFm)) + ! ((SmFm) ∗ (SmFm) − (gFm) ∗ (gFm)) = ! (g ∗ g − Sm ∗ Sm) + ! (Sm ∗ Sm − (Sm ∗ Sm)(Fm ∗ Fm)) + ! ((SmFm) ∗ (SmFm) − (gFm) ∗ (gFm)), using (4.10) for the last equality. We consider each of the three terms separately. For differentiable functions F and G we have ′ ! (F ∗ F − G ∗ G) = ! ((F + G) ∗ (F − G)) ≤ ∥((F + G) ∗ (F − G)) ∥∞ ′ ′ = ∥(F + G) ∗ (F − G) ∥∞ ≤ ∥F + G∥1∥(F − G) ∥∞ ′ ≤ (∥F ∥1 + ∥G∥1)∥(F − G) ∥∞. (4.14) For the first term, we use (4.14) to get ′ −(2k+2) ! (g ∗ g − Sm ∗ Sm) ≤ (∥g∥1 + ∥Sm∥1)∥(g − Sm) ∥∞ ≤ (2 + ∥g∥∞)2Cm by (4.8) and (4.9). Since (2 + ∥g∥∞)2C is a constant, we can take sufficiently large m that " ! (g ∗ g − S ∗ S ) < . (4.15) m m 24 For the second term, ! (Sm ∗ Sm − (Sm ∗ Sm)(Fm ∗ Fm)) = ! ((Sm ∗ Sm)(1 − Fm ∗ Fm)) ≤ ! (Sm ∗ Sm)∥1 − Fm ∗ Fm∥∞ + ! (1 − Fm ∗ Fm)∥Sm ∗ Sm∥∞ ≤ ! (Sm ∗ Sm)∥1 − Fm ∗ Fm∥∞ + ! (Fm ∗ Fm)∥Sm ∗ Sm∥∞ ≤ ! (Sm ∗ Sm) + ∥Sm ∗ Sm∥∞, 52 where the first inequality follows by Λ -multiplication, the second follows by Λ -translation invariance, and the final inequality is a consequence of (i)m and (viii)m. To handle this expression, the proof of (4.15) gives ! (Sm ∗ Sm) ≤ ! (g ∗ g) + 1 and the proof of (4.11) gives 2 ∥Sm ∗ Sm∥∞ ≤ ∥g∥∞ + 1. Thus, " ! (S ∗ S − (S ∗ S )(F ∗ F )) ≤ (! (g ∗ g) + ∥g∥2 + 2) < , (4.16) m m m m m m ∞ 24 by our choice of to satisfy (4.6). For the third term, we use (4.14) to get ! ((SmFm) ∗ (SmFm) − (gFm) ∗ (gFm)) ′ ≤ (∥SmFm∥1 + ∥gFm∥1)∥(SmFm − gFm) ∥∞ ′ ≤ (∥Sm∥∞∥Fm∥1 + ∥g∥∞∥Fm∥1)∥((Sm − g)Fm) ∥∞ ′ = (1 + 2∥g∥∞)∥((Sm − g)Fm) ∥∞, by (4.8) and (v)m. Now by the product rule, ′ ′ ′ ∥((Sm − g)Fm) ∥∞ ≤ ∥(Sm − g) Fm∥∞ + ∥(Sm − g)Fm∥∞ ′ ′ ≤ ∥(Sm − g) ∥∞∥Fm∥∞ + ∥Sm − g∥∞∥Fm∥∞ ≤ 2Cm−(2k+2)4kmk + Cm−(2k+2)42k+1m2k+1 ≤ Cm−(2k+2)42k+1m2k+1 + Cm−(2k+2)42k+1m2k+1 = 2 ⋅ 42k+1Cm−1, where we have used (4.7), (4.9), (iii)m, and (iv)m in the third inequality. Thus, 2k+1 −1 ! ((SmFm) ∗ (SmFm) − (gFm) ∗ (gFm)) ≤ 2(1 + 2∥g∥∞)4 Cm . 53 Then we can take sufficiently large m that " ! ((S F ) ∗ (S F ) − (gF ) ∗ (gF )) < . (4.17) m m m m m m 24 Equations (4.15), (4.16) and (4.17) together give " ! (g ∗ g − G ∗ G ) < . m m 8 Combining (4.4) with Claims 4.10 through 4.15, we get d ((E, f), (H, g)) = dℱ (E,H) + sup ∣fb(r) − gb(r)∣ + ∥f ∗ f − g ∗ g∥∞ + ! (f ∗ f − g ∗ g) r∈ℤ " " " " " " " < + + + + + + = ", 4 8 8 8 8 8 8 as required. Lemma 4.16. Let 1 > > 1/2 and suppose that is a generalized -power function. Then ℋn is dense in (ℳ , d ) for every n. Proof. Fix " > 0 and let (E, ) ∈ ℳ . By definition of ℳ , we can find an infinitely differentiable function g : T → ℝ and a closed set H such that (H, g) ∈ ℳ and " d ((E, ), (H, g)) < . (4.18) 2 Notice R g(t) dt = 1 since g is a probability measure. By Lemma 4.9, there exists a T positive, infinitely differentiable function f : → such that R f(t) dt = 1 and a closed T ℝ T set F ⊇ supp f such that (F, f) ∈ ℋn and " d ((F, f), (H, g)) < . (4.19) 2 Combining (4.18) and (4.19) yields the desired result. 54 Theorem 4.17. Let 1 > > 1/2 and suppose that is a generalized -power function. The complement of the set ℋ = {(E, ) ∈ ℳ : dimH (E) ≤ } is of first category in (ℳ , d ). In particular, ℋ is dense in (ℳ , d ). Proof. We claim ℋn is open in (ℳ , d ). Suppose (E, ) ∈ ℋn. By definition of ℋn, we can find a finite collection of closed intervals ℐ with [ X 1 I ⊇ E and ∣I∣ +1/n < . n I∈ℐ I∈ℐ Since ℐ is finite, we can find an > 0 such that X 1 (∣I∣ + 2) +1/n < . n I∈ℐ Define ℐe = {[a − , b + ]:[a, b] ∈ ℐ}. If (F, ) ∈ ℳ with d ((E, ), (F, )) < , then automatically, [ X 1 I ⊇ F and ∣I∣ +1/n < , n I∈ℐe I∈ℐe and so (F, ) ∈ ℋn. Thus ℋn is open in (ℳ , d ). Lemma 4.16 tells us that ℋn is dense in (ℳ , d ), so it follows that the complement of T∞ ℋn is nowhere dense, being closed. Thus, the complement of n=1 ℋn is of first category in (ℳ , d ). T∞ Suppose (E, ) ∈ n=1 ℋn, then for each n we can find a finite collection ℐ of closed intervals such that [ X 1 I ⊇ E and ∣I∣ +1/n < . n I∈ℐ I∈ℐ Assume dimH (E) > . Then there is an integer N such that dimH (E) > + 1/N. Fix > 0 and take n ≥ N sufficiently large that + 1/n < 1 and 1/n < . Then X 1 ∣I∣ ≤ ∣I∣ +1/n ≤ ∣I∣ +1/n < < , n I∈ℐ 55 S and I∈ℐ I ⊇ E. In particular, this gives a uniform bound +1/N +1/n X 1 ℋ (E) ≤ ℋ (E) ≤ ∣I∣ +1/n < ≤ 1 < ∞, n I∈ℐ for every > 0. Taking the supremum over > 0, we get that ℋ +1/N (E) ≤ 1 < ∞, and so dimH (E) ≤ + 1/N, a contradiction. Therefore, dimH (E) ≤ , and so (E, ) ∈ ℋ. c T∞ c This gives ℋ ⊆ ( n=1 ℋn) . Since any subset of a set of first category is of first category, c S∞ this completes the first part of the proof. We will suppose ℋ = n=1 En for some nowhere dense sets En. By Lemma 4.8,(ℳ , d ) is a complete metric space, so the Baire Category theorem tells us that this space is Baire, that is, the countable intersection of open dense sets is dense. T∞ c The interior of the complement of a nowhere dense set is dense, so that n=1 int(En) is dense as ℳ is Baire. Then ℋ contains a dense set, and thus is itself dense in (ℳ , d ). We finally arrive at a somewhat generalized form of our main theorem: Theorem 4.18. Let 1 > > 1/2 and be a generalized -power function. Then there exists a probability measure such that dimH (supp ) ≤ and ∗ = f with f ∈ Λ . Proof. By Theorem 4.17, ℋ is non-empty. Thus we can find (E, ) ∈ ℳ with dimH (E) ≤ , and so dimH (supp ) ≤ . By definition of ℳ , we also have that is a probability measure with ∗ = f with f ∈ Λ . Theorem 4.19. If 1 > > 1/2, then there exists a probability measure such that dimH (supp ) = and ∗ = f where f ∈ Λ −1/2. −1/2 Proof. Taking (t) = t , Theorem 4.18 gives us a probability measure with dimH (supp ) ≤ and ∗ = f with f ∈ Λ = Λ −1/2. By Lemma 1.11, 1 1 dim (supp ) − ≥ − , H 2 2 and hence dimH (supp ) = . 56 APPENDICES 57 Appendix A Measure Theory Some of these results are applicable in the more general case of -finite measures, but for our purposes, we will only require finite measures. As such, we will make the assumption that all measures are finite. Measures will also be taken to be Borel and regular. Definition A.1. Suppose (Ω, Σ) is a measure space and and are complex measures on (Ω, Σ). is said to be absolutely continuous with respect if whenever (E) = 0 for a set E ∈ Σ, then we have (E) = 0. In this case, we typically write ≪ . If Ω is a topological group and is the Haar measure, we typically say is absolutely continuous. Absolutely continuous measures are nice, due to their correspondence with L1 functions. Theorem A.2 [Radon-Nikodym theorem]. Suppose (Ω, Σ) is a measure space, and , are measures on (Ω, Σ). Then ≪ if and only if there exists a function f ∈ L1() such that Z (E) = f d E for any measurable set E ∈ Σ. f is referred to as the Radon-Nikodym derivative. Such a correspondence allows us to treat absolutely continuous measures as functions, by taking their Radon-Nikodym derivatives. 59 Definition A.3. Suppose (Ω, Σ) is a measure space. A measure on (Ω, Σ) is said to be singular with respect to a measure on (Ω, Σ) if there exist two disjoint sets E,F ∈ Σ with E ∪F = Ω such that is zero on every subset of E and is zero on every subset of F . In this case, we typically write ⊥ . As this definition is symmetric in and , we may sometimes simply say and are singular. Notice that in the case of positive measures, it suffices to have the condition (E) = 0 = (F ). If Ω is a topological group and is the Haar measure, we typically say is singular. Theorem A.4 [Lebesgue’s Decomposition theorem]. Suppose (Ω, Σ) is a measure space, and , are measures on (Ω, Σ). Then there exist two measures ac and s such that (i) = ac + s, (ii) ac ≪ , and (iii) s ⊥ . Furthermore, these two measures are uniquely determined. Both Theorems A.2 and A.4 are standard results that can be found in any good measure theory (see, for example, [13] Theorem 6.9). In the special case of being Lebesgue measure, Lebesgue’s Decomposition theorem may be further extended to include the concepts of continuous measures and discrete measures. Definition A.5. Suppose (Ω, Σ) is a measure space. A measure on (Ω, Σ) is said to be continuous if ({x}) = 0 for every x ∈ Ω. A measure is said to be discrete if it is concentrated on a countable set, that is, if it can be expressed as ∞ X = anxn , n=1 60 where an are constants, xn ∈ Ω, and x is the Dirac measure defined by ⎧ ⎨1 if x ∈ E, x(E) = ⎩0 if x∈ / E. Recall that we assume measures are finite, so this summation is finite. The next lemma shows another decomposition of the space of measures, this time using continuous and discrete measures. Lemma A.6. Suppose (Ω, Σ) is a measure space, and is a measure on (Ω, Σ). Then there exist two measures c and d such that (i) = c + d, (ii) c is continuous, and (iii) d is discrete. Furthermore, these two measures are uniquely determined. P Proof. Let E = {x : ({x}) ∕= 0}. For each finite subset F ⊆ E, x∈F ∣({x})∣ ≤ ∥∥M(G) < ∞. Taking the supremum over all possible finite subsets F ⊆ F , we get P x∈E ∣({x})∣ ≤ ∥∥M(G) < ∞. Thus for every n ∈ ℕ, there can only be finitely many x ∈ E such that ({x}) ≥ 1/n, and so E must be countable, say E = {x1, x2,...}. Define P∞ the discrete measure d = n=1 ({xn})xn , and define c = −d. Then for every x ∈ Ω, ⎧ ⎨0 if x∈ / E, c({x}) = ⎩({xn}) − d({xn}) = 0 if x = xn ∈ E, so that c is continuous. ′ ′ For uniqueness, suppose c is a continuous measure and d is a discrete measure, ′ ′ ′ ′ ′ ′ with = c + d. Then c + d = c + d yields c − c = d − d. It is clear from the definitions that both the continuous measures and discrete measures are closed under linear 61 ′ combinations, so we have the continuous measure c − c equal to the discrete measure ′ P d − d = n anyn . Since this is continuous, for each n the discrete measure must assign ′ ′ a weight of 0 to the point yn, which implies an = 0. Therefore c = c and d = d, so the decomposition is unique. It is clear from the definition of the support of a measure and a discrete measure that P∞ if d = n=1 anxn is a discrete measure, then supp d = {xn : an ∕= 0}. In particular, d is supported on a countable set, and hence is easily seen to be singular with respect to Lebesgue measure. Then by Lemma A.6, the singular measures can be decomposed into a discrete part and a part that is both singular and continuous, which we call continuous singular. This proves the following extension of Lebesgue’s Decomposition theorem: Theorem A.7. Suppose is a -finite measure on ℝn or Tn. Then there exist three -finite measures ac, cs, and d such that (i) = ac + cs + d, (ii) ac is absolutely continuous, (iii) cs is continuous singular, and (iv) d is discrete. Furthermore, these three measures are uniquely determined. 62 Appendix B Convergence of Fourier Series Definition B.1. Suppose we have two functions, f(x) and g(x). We write f(x) = o(g(x)) as x → 0 if and only if f(x)/g(x) → 0 as x → 0. For brevity, we typically omit the condition x → 0 and simply write f(x) = o(g(x)). Theorem B.2 [Dini-Lipschitz test]. As a function of ℎ, if f(x + ℎ) − f(x) = o (ln ∣ℎ∣−1)−1 uniformly in x on T, then the Fourier series of f converges uniformly on T. For a proof and more detailed discussion, refer to [1] Chapter IV §4. 63 Appendix C Probability Theory Theorem C.1 [Markov’s inequality]. If X is a non-negative random variable, then for all > 0, P(X ≥ ) ≤ E(X) −1. Proof. Define a random variable Y by ⎧ ⎨ if X(x) ≥ , Y (x) = ⎩0 otherwise. Clearly Y ≤ X, so that E(Y ) ≤ E(X) by monotonicity. On the other hand, we have by direct computation that E(Y ) = P(X ≥ ). The result follows immediately. 65 Appendix D The Hausdorff Metric Definition D.1. Suppose (X, d) is a metric space, and let ℱ be the family of all closed and bounded subsets of X. For two sets E,F ∈ ℱ, define dℱ (E,F ) = max sup inf d(e, f), sup inf d(e, f) . e∈E f∈F f∈F e∈E We call dℱ the Hausdorff metric. It is worth noting that there is a natural reason to restrict to sets belonging to ℱ. Indeed, if we allow arbitrary sets to be measured in this way, then we shall see dℱ (E,F ) = 0 if and only if E and F have the same closure, which does not necessarily mean E = F , violating the definition for a metric. As well, allowing unbounded sets will allow dℱ to take on the value ∞, which is undesirable. We now show that the term “metric” is used here in a natural way, that is, we do in fact have a metric. Theorem D.2. (ℱ, dℱ ) is a metric space. Proof. Clearly if E,F ∈ ℱ with E = F , then dℱ (E,F ) = 0. Conversely, if dℱ (E,F ) = 0 then sup inf d(e, f) = 0 = sup inf d(e, f). e∈E f∈F f∈F e∈E 67 Since these infima are non-negative, we get inf d(e, f) = 0 = inf d(e, f). f∈F e∈E As E and F are closed, this implies E = F . Trivially, dℱ (E,F ) = dℱ (F,E) for any E,F ∈ ℱ. Finally, suppose E,F,G ∈ ℱ and choose e0 ∈ E, f0 ∈ F . By triangle inequality, inf d(e0, g) ≤ inf (d(e0, f0) + d(f0, g)) g∈G g∈G = d(e0, f0) + inf d(f0, g) g∈G ≤ d(e0, f0) + sup inf d(f, g). f∈F g∈G Since this holds for any f0 ∈ F , it holds in the infimum, inf d(e0, g) ≤ inf d(e0, f) + sup inf d(f, g) g∈G f∈F f∈F g∈G ≤ sup inf d(e, f) + sup inf d(f, g), e∈E f∈F f∈F g∈G so that taking supremum, sup inf d(e, g) ≤ sup inf d(e, f) + sup inf d(f, g). e∈E g∈G e∈E f∈F f∈F g∈G By symmetry, we get sup inf d(g, e) ≤ sup inf d(g, f) + sup inf d(f, e). g∈G e∈E g∈G f∈F f∈F e∈E Combining these equations together gives dℱ (E,G) ≤ dℱ (E,F ) + dℱ (F,G). Thus, (ℱ, dℱ ) is indeed a metric space. A nice property of the Hausdorff metric is that it inherits completeness from the un- derlying metric. 68 Theorem D.3. If (X, d) is a complete metric space, then (ℱ, dℱ ) is a complete metric space. Proof. Theorem D.2 ensures (ℱ, dℱ ) is a metric space. To see completeness, let E1,E2,... be a Cauchy sequence in (ℱ, dℱ ). By taking a subsequence if necessary, we may assume −n−1 dℱ (En,En+1) < 2 . Define −n E = {e = lim en : en ∈ En and d(en, en+1) < 2 }. n→∞ To see E is non-empty, we will inductively create a sequence satisfying the necessary property for its limit to lie in E. Since E1 ∈ ℱ, it is non-empty, thus we can choose e1 ∈ E1. Inductively, having chosen en ∈ En, our assumption gives ( ) inf d(en, f) ≤ max sup inf d(g, f), sup inf d(g, f) f∈E f∈E g∈E n+1 g∈En n+1 f∈En+1 n −n−1 = dℱ (En,En+1) < 2 . Since En+1 is closed, the infimum is attained and so there exists en+1 ∈ En+1 such that −n−1 d(en, en+1) = inf d(en, f) < 2 . f∈En+1 Thus {en} forms a Cauchy sequence. Since X is complete with respect to d, en converges to some e ∈ X. That is, e ∈ E, so E is non-empty. Notice that by triangle inequality, we also get that for this e, −n+1 d(en, e) < 2 . Let " > 0. Let N be sufficiently large that 2−N+1 < ". Suppose n ≥ N and take −n+1 en ∈ En. By the construction above, we can get e ∈ E with d(en, e) < 2 . Then certainly −n+1 inf d(en, f) < 2 . f∈E Since this holds for any en ∈ En, sup inf d(g, f) ≤ 2−n+1. g∈En f∈E 69 Conversely, take e ∈ E. By definition, there is a sequence en with e = limn→∞ en, en ∈ En, −n −n+1 and d(en, en+1) < 2 . Then by triangle inequality, d(en, e) < 2 , and so certainly inf d(g, e) < 2−n+1. g∈En Since this holds for any e ∈ E, sup inf d(g, f) ≤ 2−n+1. f∈E g∈En The suprema and infima are unchanged if we replace E with its closure E. Then together, these give ( ) −n+1 −N+1 dℱ (En, E) = max sup inf d(g, f), sup inf d(g, f) ≤ 2 ≤ 2 < ". g∈E g∈En f∈E f∈E n Hence, En converges to E ∈ ℱ in the dℱ metric, that is, (ℱ, dℱ ) is a complete metric space. Two sets are close in the Hausdorff metric if they lie in a “thickened” version of the other set. To be precise, let (X, d) be a metric space. 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